Closed form for $I(b)=int_0^1arctan^bx mathrm{d}x$











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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







share|cite|improve this question




















  • 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















up vote
3
down vote

favorite
3












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







share|cite|improve this question




















  • 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













up vote
3
down vote

favorite
3









up vote
3
down vote

favorite
3






3





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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




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edited Nov 23 at 21:52

























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














  • 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










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}






share|cite|improve this answer























  • 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













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1 Answer
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1 Answer
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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}






share|cite|improve this answer























  • 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

















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}






share|cite|improve this answer























  • 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















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}






share|cite|improve this answer














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}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 7:13

























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




















  • 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




















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