Closed form for $I(b)=int_0^1arctan^bx mathrm{d}x$
up vote
3
down vote
favorite
Just out of curiosity, I am trying to find a closed form for
$$I(b)=int_0^1arctan^bt mathrm{d}t$$
It converges for real $b>-1$, and behaves a lot like $frac1{1+x}$.
I started out with noting that $$arctan x=frac i2logfrac{1-ix}{1+ix}$$
$$arctan^bx=frac{i^b}{2^b}log^bfrac{1-ix}{1+ix}$$
Which results in $$I(b)=frac{i^b}{2^b}int_0^1log^bfrac{1-it}{1+it}mathrm{d}t$$
Which I do not know how to find.
I did not want to give up, so I found
$$I(1)=int_0^1arctan t mathrm{d}t$$
Using $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-(Fcirc f^{-1})(x)$$
We have that
$$I(1)=fracpi4-frac12log2$$
Also, rather trivially, it is easily shown that
$$I(0)=1$$
But I still want to know if there is a closed form for $I(b)$. Can anyone help?
Update:
With the substitution $tan u=t$, we the integral becomes
$$I(b)=int_0^{fracpi4}u^bsec^2u mathrm{d}u$$
Then we note that, for $|x|<frac{pi}2$,
$$tan x=sum_{ngeq1}frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}$$
We differentiate both sides to obtain
$$sec^2x=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-2}$$
And because $[0,fracpi4]subset(frac{-pi}2,frac{pi}2)$, we have that
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}int_0^{fracpi4}u^{2n+b-2}mathrm{d}u$$
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi4bigg)^{2n+b-1}$$
$$I(b)=bigg(fracpi4bigg)^{b-1}sum_{ngeq1}frac{(-1)^{n-1}(2n-1)(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi2bigg)^{2n}$$
So, do any of you series-wizards out there have any suggestions?
integration definite-integrals
asked Nov 20 at 6:41
clathratus
2,300322
add a comment |
up vote
3
down vote
favorite
Just out of curiosity, I am trying to find a closed form for
$$I(b)=int_0^1arctan^bt mathrm{d}t$$
It converges for real $b>-1$, and behaves a lot like $frac1{1+x}$.
I started out with noting that $$arctan x=frac i2logfrac{1-ix}{1+ix}$$
$$arctan^bx=frac{i^b}{2^b}log^bfrac{1-ix}{1+ix}$$
Which results in $$I(b)=frac{i^b}{2^b}int_0^1log^bfrac{1-it}{1+it}mathrm{d}t$$
Which I do not know how to find.
I did not want to give up, so I found
$$I(1)=int_0^1arctan t mathrm{d}t$$
Using $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-(Fcirc f^{-1})(x)$$
We have that
$$I(1)=fracpi4-frac12log2$$
Also, rather trivially, it is easily shown that
$$I(0)=1$$
But I still want to know if there is a closed form for $I(b)$. Can anyone help?
Update:
With the substitution $tan u=t$, we the integral becomes
$$I(b)=int_0^{fracpi4}u^bsec^2u mathrm{d}u$$
Then we note that, for $|x|<frac{pi}2$,
$$tan x=sum_{ngeq1}frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}$$
We differentiate both sides to obtain
$$sec^2x=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-2}$$
And because $[0,fracpi4]subset(frac{-pi}2,frac{pi}2)$, we have that
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}int_0^{fracpi4}u^{2n+b-2}mathrm{d}u$$
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi4bigg)^{2n+b-1}$$
$$I(b)=bigg(fracpi4bigg)^{b-1}sum_{ngeq1}frac{(-1)^{n-1}(2n-1)(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi2bigg)^{2n}$$
So, do any of you series-wizards out there have any suggestions?
integration definite-integrals
asked Nov 20 at 6:41
clathratus
2,300322
1
Interesting problem, for sure ! I bet that $zeta(.)$ and polygamma functions would appear somewhere.
– Claude Leibovici
Nov 20 at 9:26
1
I'm betting on hypergeometric functions =))))
– Mikalai Parshutsich
Nov 20 at 10:00
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Just out of curiosity, I am trying to find a closed form for
$$I(b)=int_0^1arctan^bt mathrm{d}t$$
It converges for real $b>-1$, and behaves a lot like $frac1{1+x}$.
I started out with noting that $$arctan x=frac i2logfrac{1-ix}{1+ix}$$
$$arctan^bx=frac{i^b}{2^b}log^bfrac{1-ix}{1+ix}$$
Which results in $$I(b)=frac{i^b}{2^b}int_0^1log^bfrac{1-it}{1+it}mathrm{d}t$$
Which I do not know how to find.
I did not want to give up, so I found
$$I(1)=int_0^1arctan t mathrm{d}t$$
Using $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-(Fcirc f^{-1})(x)$$
We have that
$$I(1)=fracpi4-frac12log2$$
Also, rather trivially, it is easily shown that
$$I(0)=1$$
But I still want to know if there is a closed form for $I(b)$. Can anyone help?
Update:
With the substitution $tan u=t$, we the integral becomes
$$I(b)=int_0^{fracpi4}u^bsec^2u mathrm{d}u$$
Then we note that, for $|x|<frac{pi}2$,
$$tan x=sum_{ngeq1}frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}$$
We differentiate both sides to obtain
$$sec^2x=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-2}$$
And because $[0,fracpi4]subset(frac{-pi}2,frac{pi}2)$, we have that
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}int_0^{fracpi4}u^{2n+b-2}mathrm{d}u$$
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi4bigg)^{2n+b-1}$$
$$I(b)=bigg(fracpi4bigg)^{b-1}sum_{ngeq1}frac{(-1)^{n-1}(2n-1)(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi2bigg)^{2n}$$
So, do any of you series-wizards out there have any suggestions?
integration definite-integrals
asked Nov 20 at 6:41
clathratus
2,300322
Just out of curiosity, I am trying to find a closed form for
$$I(b)=int_0^1arctan^bt mathrm{d}t$$
It converges for real $b>-1$, and behaves a lot like $frac1{1+x}$.
I started out with noting that $$arctan x=frac i2logfrac{1-ix}{1+ix}$$
$$arctan^bx=frac{i^b}{2^b}log^bfrac{1-ix}{1+ix}$$
Which results in $$I(b)=frac{i^b}{2^b}int_0^1log^bfrac{1-it}{1+it}mathrm{d}t$$
Which I do not know how to find.
I did not want to give up, so I found
$$I(1)=int_0^1arctan t mathrm{d}t$$
Using $$int f^{-1}(x)mathrm{d}x=xf^{-1}(x)-(Fcirc f^{-1})(x)$$
We have that
$$I(1)=fracpi4-frac12log2$$
Also, rather trivially, it is easily shown that
$$I(0)=1$$
But I still want to know if there is a closed form for $I(b)$. Can anyone help?
Update:
With the substitution $tan u=t$, we the integral becomes
$$I(b)=int_0^{fracpi4}u^bsec^2u mathrm{d}u$$
Then we note that, for $|x|<frac{pi}2$,
$$tan x=sum_{ngeq1}frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}$$
We differentiate both sides to obtain
$$sec^2x=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-2}$$
And because $[0,fracpi4]subset(frac{-pi}2,frac{pi}2)$, we have that
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}int_0^{fracpi4}u^{2n+b-2}mathrm{d}u$$
$$I(b)=sum_{ngeq1}frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi4bigg)^{2n+b-1}$$
$$I(b)=bigg(fracpi4bigg)^{b-1}sum_{ngeq1}frac{(-1)^{n-1}(2n-1)(4^n-1)B_{2n}}{(2n)!(2n+b-1)}bigg(fracpi2bigg)^{2n}$$
So, do any of you series-wizards out there have any suggestions?
integration definite-integrals
integration definite-integrals
asked Nov 20 at 6:41
clathratus
2,300322
asked Nov 20 at 6:41
clathratus
2,300322
edited Nov 23 at 21:52
asked Nov 20 at 6:41
clathratus
2,300322
asked Nov 20 at 6:41
clathratus
2,300322
asked Nov 20 at 6:41
clathratus
2,300322
2,300322
1
Interesting problem, for sure ! I bet that $zeta(.)$ and polygamma functions would appear somewhere.
– Claude Leibovici
Nov 20 at 9:26
1
I'm betting on hypergeometric functions =))))
– Mikalai Parshutsich
Nov 20 at 10:00
add a comment |
1
Interesting problem, for sure ! I bet that $zeta(.)$ and polygamma functions would appear somewhere.
– Claude Leibovici
Nov 20 at 9:26
1
I'm betting on hypergeometric functions =))))
– Mikalai Parshutsich
Nov 20 at 10:00
1
1
Interesting problem, for sure ! I bet that $zeta(.)$ and polygamma functions would appear somewhere.
– Claude Leibovici
Nov 20 at 9:26
Interesting problem, for sure ! I bet that $zeta(.)$ and polygamma functions would appear somewhere.
– Claude Leibovici
Nov 20 at 9:26
1
1
I'm betting on hypergeometric functions =))))
– Mikalai Parshutsich
Nov 20 at 10:00
I'm betting on hypergeometric functions =))))
– Mikalai Parshutsich
Nov 20 at 10:00
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Here is an attempt that is too long for a comment. Maybe you'll find it useful.
begin{align}
int_0^1 arctan ^b x,dx &= frac{i^b}{2^b} int_0^1 log^b left(frac{1-ix}{1+ix} right) \
&= frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(frac{1-ix}{1+ix} right)^alpha \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(1-ixright)^{2alpha}(1+x^2)^{-alpha} ,dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^1 left(1-ixright)^{2alpha}int_0^infty nu^{alpha-1} e^{-nu(1+x^2)} ,dnu dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 left(1-ixright)^{2alpha} e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k int_0^1 x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2},nu right) right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2}right) e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty e^{-nu} nu^{alpha-k/2-3/2} left( 1 - e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty left( e^{-nu} nu^{alpha-k/2-3/2} - sum_{l=0}^{k/2-1/2} frac{e^{-2nu} nu^{alpha-k/2-3/2+l}}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( Gammaleft( alpha-k/2-1/2 right) - sum_{l=0}^{k/2-1/2} frac{Gamma left( alpha-k/2-1/2+lright) }{2^{alpha-k/2-1/2+l}Gamma(l+1)}right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)Gamma(alpha)}{2^alpha Gamma(k/2+3/2)} right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k ,frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)}{2^alpha Gamma(k/2+3/2)} \
end{align}
answered Nov 21 at 4:31
Zachary
2,2141211
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Here is an attempt that is too long for a comment. Maybe you'll find it useful.
begin{align}
int_0^1 arctan ^b x,dx &= frac{i^b}{2^b} int_0^1 log^b left(frac{1-ix}{1+ix} right) \
&= frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(frac{1-ix}{1+ix} right)^alpha \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(1-ixright)^{2alpha}(1+x^2)^{-alpha} ,dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^1 left(1-ixright)^{2alpha}int_0^infty nu^{alpha-1} e^{-nu(1+x^2)} ,dnu dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 left(1-ixright)^{2alpha} e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k int_0^1 x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2},nu right) right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2}right) e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty e^{-nu} nu^{alpha-k/2-3/2} left( 1 - e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty left( e^{-nu} nu^{alpha-k/2-3/2} - sum_{l=0}^{k/2-1/2} frac{e^{-2nu} nu^{alpha-k/2-3/2+l}}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( Gammaleft( alpha-k/2-1/2 right) - sum_{l=0}^{k/2-1/2} frac{Gamma left( alpha-k/2-1/2+lright) }{2^{alpha-k/2-1/2+l}Gamma(l+1)}right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)Gamma(alpha)}{2^alpha Gamma(k/2+3/2)} right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k ,frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)}{2^alpha Gamma(k/2+3/2)} \
end{align}
answered Nov 21 at 4:31
Zachary
2,2141211
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
add a comment |
up vote
4
down vote
accepted
Here is an attempt that is too long for a comment. Maybe you'll find it useful.
begin{align}
int_0^1 arctan ^b x,dx &= frac{i^b}{2^b} int_0^1 log^b left(frac{1-ix}{1+ix} right) \
&= frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(frac{1-ix}{1+ix} right)^alpha \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(1-ixright)^{2alpha}(1+x^2)^{-alpha} ,dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^1 left(1-ixright)^{2alpha}int_0^infty nu^{alpha-1} e^{-nu(1+x^2)} ,dnu dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 left(1-ixright)^{2alpha} e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k int_0^1 x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2},nu right) right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2}right) e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty e^{-nu} nu^{alpha-k/2-3/2} left( 1 - e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty left( e^{-nu} nu^{alpha-k/2-3/2} - sum_{l=0}^{k/2-1/2} frac{e^{-2nu} nu^{alpha-k/2-3/2+l}}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( Gammaleft( alpha-k/2-1/2 right) - sum_{l=0}^{k/2-1/2} frac{Gamma left( alpha-k/2-1/2+lright) }{2^{alpha-k/2-1/2+l}Gamma(l+1)}right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)Gamma(alpha)}{2^alpha Gamma(k/2+3/2)} right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k ,frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)}{2^alpha Gamma(k/2+3/2)} \
end{align}
answered Nov 21 at 4:31
Zachary
2,2141211
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Here is an attempt that is too long for a comment. Maybe you'll find it useful.
begin{align}
int_0^1 arctan ^b x,dx &= frac{i^b}{2^b} int_0^1 log^b left(frac{1-ix}{1+ix} right) \
&= frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(frac{1-ix}{1+ix} right)^alpha \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(1-ixright)^{2alpha}(1+x^2)^{-alpha} ,dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^1 left(1-ixright)^{2alpha}int_0^infty nu^{alpha-1} e^{-nu(1+x^2)} ,dnu dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 left(1-ixright)^{2alpha} e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k int_0^1 x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2},nu right) right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2}right) e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty e^{-nu} nu^{alpha-k/2-3/2} left( 1 - e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty left( e^{-nu} nu^{alpha-k/2-3/2} - sum_{l=0}^{k/2-1/2} frac{e^{-2nu} nu^{alpha-k/2-3/2+l}}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( Gammaleft( alpha-k/2-1/2 right) - sum_{l=0}^{k/2-1/2} frac{Gamma left( alpha-k/2-1/2+lright) }{2^{alpha-k/2-1/2+l}Gamma(l+1)}right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)Gamma(alpha)}{2^alpha Gamma(k/2+3/2)} right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k ,frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)}{2^alpha Gamma(k/2+3/2)} \
end{align}
answered Nov 21 at 4:31
Zachary
2,2141211
Here is an attempt that is too long for a comment. Maybe you'll find it useful.
begin{align}
int_0^1 arctan ^b x,dx &= frac{i^b}{2^b} int_0^1 log^b left(frac{1-ix}{1+ix} right) \
&= frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(frac{1-ix}{1+ix} right)^alpha \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0}int_0^1 left(1-ixright)^{2alpha}(1+x^2)^{-alpha} ,dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^1 left(1-ixright)^{2alpha}int_0^infty nu^{alpha-1} e^{-nu(1+x^2)} ,dnu dx \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 left(1-ixright)^{2alpha} e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} int_0^1 sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^b} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-1} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k int_0^1 x^k e^{-nu x^2} ,dxdnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2},nu right) right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)}int_0^infty e^{-nu} nu^{alpha-3/2} sum_{k=0}^{2alpha} {2alphachoose k} (-1)^k i^k nu^{-k/2} left( Gammaleft( frac{k+1}{2}right) - Gammaleft( frac{k+1}{2}right) e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty e^{-nu} nu^{alpha-k/2-3/2} left( 1 - e^{-nu} sum_{l=0}^{k/2-1/2} frac{nu^l}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k int_0^infty left( e^{-nu} nu^{alpha-k/2-3/2} - sum_{l=0}^{k/2-1/2} frac{e^{-2nu} nu^{alpha-k/2-3/2+l}}{Gamma(l+1)}right) dnu \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( Gammaleft( alpha-k/2-1/2 right) - sum_{l=0}^{k/2-1/2} frac{Gamma left( alpha-k/2-1/2+lright) }{2^{alpha-k/2-1/2+l}Gamma(l+1)}right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} frac{1}{Gamma(alpha)} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k left( frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)Gamma(alpha)}{2^alpha Gamma(k/2+3/2)} right) \
&=frac{i^b}{2^{b+1}} frac{partial^b}{partial alpha^b}Biggr|_{alpha=0} sum_{k=0}^{2alpha} Gammaleft( frac{k+1}{2}right){2alphachoose k} (-1)^k i^k ,frac{_2F_1left(1,alpha; k/2+3/2;1/2 right)}{2^alpha Gamma(k/2+3/2)} \
end{align}
answered Nov 21 at 4:31
Zachary
2,2141211
edited Nov 21 at 7:13
answered Nov 21 at 4:31
Zachary
2,2141211
answered Nov 21 at 4:31
Zachary
2,2141211
answered Nov 21 at 4:31
Zachary
2,2141211
2,2141211
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
add a comment |
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you.
– marty cohen
Nov 21 at 4:42
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
How is it now haha? I did manage to integrate everything out.
– Zachary
Nov 21 at 4:45
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Wow that's pretty amazing. I'll have to examine it before I accept it.
– clathratus
Nov 21 at 6:33
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
Looks legit. Does this only work for $binBbb N$?
– clathratus
Nov 21 at 18:12
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
This method actually works for any $bin(-1,infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order.
– Zachary
Nov 21 at 21:08
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006029%2fclosed-form-for-ib-int-01-arctanbx-mathrmdx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Interesting problem, for sure ! I bet that $zeta(.)$ and polygamma functions would appear somewhere.
– Claude Leibovici
Nov 20 at 9:26
1
I'm betting on hypergeometric functions =))))
– Mikalai Parshutsich
Nov 20 at 10:00