Substitution in an Integral from log to inverse
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Considering equations (2) (8) and (9) below. Please, I am wondering what is the insight into how the substitution was made to get (9). I have exhausted much energy trying to figure this out.
Note that the (2) in red is what I think was substituted into (8) from my own understanding. Also, how the $log (1+v)$ was changed to $frac{1}{v+1}$ is not clear to me. Pardon this seemingly easy but deep question.
integration substitution
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
$endgroup$
add a comment |
$begingroup$
Considering equations (2) (8) and (9) below. Please, I am wondering what is the insight into how the substitution was made to get (9). I have exhausted much energy trying to figure this out.
Note that the (2) in red is what I think was substituted into (8) from my own understanding. Also, how the $log (1+v)$ was changed to $frac{1}{v+1}$ is not clear to me. Pardon this seemingly easy but deep question.
integration substitution
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
$endgroup$
$begingroup$
This is integration by parts.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:21
add a comment |
$begingroup$
Considering equations (2) (8) and (9) below. Please, I am wondering what is the insight into how the substitution was made to get (9). I have exhausted much energy trying to figure this out.
Note that the (2) in red is what I think was substituted into (8) from my own understanding. Also, how the $log (1+v)$ was changed to $frac{1}{v+1}$ is not clear to me. Pardon this seemingly easy but deep question.
integration substitution
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
$endgroup$
Considering equations (2) (8) and (9) below. Please, I am wondering what is the insight into how the substitution was made to get (9). I have exhausted much energy trying to figure this out.
Note that the (2) in red is what I think was substituted into (8) from my own understanding. Also, how the $log (1+v)$ was changed to $frac{1}{v+1}$ is not clear to me. Pardon this seemingly easy but deep question.
integration substitution
integration substitution
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
edited Dec 28 '18 at 3:49
Abdulhameed
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
asked Dec 28 '18 at 3:10
AbdulhameedAbdulhameed
153115
153115
$begingroup$
This is integration by parts.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:21
add a comment |
$begingroup$
This is integration by parts.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:21
$begingroup$
This is integration by parts.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:21
$begingroup$
This is integration by parts.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$C_{D2D}=lambda_dfrac1Nint_0^inftylog_2(1+nu)p(nu)dnu\=lambda_dfrac1{Nln2}int_0^infty ln(1+nu)p(nu)dnu$$Integrating by parts, and letting $P(nu)=int_infty^nu p(t) dt$,
$$C_{D2D}=lambda_dfrac1{Nln2}left[ln(1+nu)P(nu)right]_0^infty-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}P(nu)dnu$$The boundary term vanishes, leaving $$C_{D2D}=-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}(-exp(-zeta_{dr}(lambda_d+gamma_{dc}lambda_c)))dnu$$
which is as required (though I am not too sure why the $ln 2$ is not included in the solution in the paper).
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
$endgroup$
2
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
1
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
|
show 2 more comments
$begingroup$
Integration by parts works as follows: $$int f'g=fg-int fg'$$.
That would explain the appearance of $$(log(1+nu))'=frac1{1+nu}$$.
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
$$C_{D2D}=lambda_dfrac1Nint_0^inftylog_2(1+nu)p(nu)dnu\=lambda_dfrac1{Nln2}int_0^infty ln(1+nu)p(nu)dnu$$Integrating by parts, and letting $P(nu)=int_infty^nu p(t) dt$,
$$C_{D2D}=lambda_dfrac1{Nln2}left[ln(1+nu)P(nu)right]_0^infty-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}P(nu)dnu$$The boundary term vanishes, leaving $$C_{D2D}=-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}(-exp(-zeta_{dr}(lambda_d+gamma_{dc}lambda_c)))dnu$$
which is as required (though I am not too sure why the $ln 2$ is not included in the solution in the paper).
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
$endgroup$
2
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
1
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
|
show 2 more comments
$begingroup$
$$C_{D2D}=lambda_dfrac1Nint_0^inftylog_2(1+nu)p(nu)dnu\=lambda_dfrac1{Nln2}int_0^infty ln(1+nu)p(nu)dnu$$Integrating by parts, and letting $P(nu)=int_infty^nu p(t) dt$,
$$C_{D2D}=lambda_dfrac1{Nln2}left[ln(1+nu)P(nu)right]_0^infty-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}P(nu)dnu$$The boundary term vanishes, leaving $$C_{D2D}=-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}(-exp(-zeta_{dr}(lambda_d+gamma_{dc}lambda_c)))dnu$$
which is as required (though I am not too sure why the $ln 2$ is not included in the solution in the paper).
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
$endgroup$
2
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
1
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
|
show 2 more comments
$begingroup$
$$C_{D2D}=lambda_dfrac1Nint_0^inftylog_2(1+nu)p(nu)dnu\=lambda_dfrac1{Nln2}int_0^infty ln(1+nu)p(nu)dnu$$Integrating by parts, and letting $P(nu)=int_infty^nu p(t) dt$,
$$C_{D2D}=lambda_dfrac1{Nln2}left[ln(1+nu)P(nu)right]_0^infty-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}P(nu)dnu$$The boundary term vanishes, leaving $$C_{D2D}=-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}(-exp(-zeta_{dr}(lambda_d+gamma_{dc}lambda_c)))dnu$$
which is as required (though I am not too sure why the $ln 2$ is not included in the solution in the paper).
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
$endgroup$
$$C_{D2D}=lambda_dfrac1Nint_0^inftylog_2(1+nu)p(nu)dnu\=lambda_dfrac1{Nln2}int_0^infty ln(1+nu)p(nu)dnu$$Integrating by parts, and letting $P(nu)=int_infty^nu p(t) dt$,
$$C_{D2D}=lambda_dfrac1{Nln2}left[ln(1+nu)P(nu)right]_0^infty-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}P(nu)dnu$$The boundary term vanishes, leaving $$C_{D2D}=-lambda_dfrac1{Nln2}int_0^infty frac1{1+nu}(-exp(-zeta_{dr}(lambda_d+gamma_{dc}lambda_c)))dnu$$
which is as required (though I am not too sure why the $ln 2$ is not included in the solution in the paper).
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
edited Dec 28 '18 at 4:55
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
answered Dec 28 '18 at 3:35
John DoeJohn Doe
11.2k11239
11.2k11239
2
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
1
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
|
show 2 more comments
2
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
1
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
2
2
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
$begingroup$
The missing $ln 2$ was also bothering me.
$endgroup$
– Randall
Dec 28 '18 at 3:44
1
1
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
Thanks John for this followable answer
$endgroup$
– Abdulhameed
Dec 28 '18 at 3:45
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@John, please at your convenience, may I see how intermediate steps between the third and fourth lines (The integration by parts) was done and the rationale behind the boundary term vanishing. I think I got something mixed up in my own trial. Thanks.
$endgroup$
– Abdulhameed
Dec 28 '18 at 4:42
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
@Abdulhameed For the integration by parts, use $$int f frac{dg}{dnu} dnu=fg-intfrac{df}{dnu}gdnu$$ with $f(nu)=ln (1+nu)$ and $frac{dg(nu)}{dnu}=p(nu)$. Then $g(nu)=P(nu)$ due to how I defined $P(nu)$ as the integral of $p(nu)$. Differentiating the log gives us the $frac1{1+nu}$ term.
$endgroup$
– John Doe
Dec 28 '18 at 4:54
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
$begingroup$
The boundary term is (corrected from before), given by $$begin{align}left[ln(1+nu)P(nu)right]_0^infty&=lim_{nutoinfty}(ln(1+nu)P(nu))-ln(1)P(0)\&=-lim_{nutoinfty}left(ln(1+nu)int_nu^infty p(x) dxright)\&=-lim_{nutoinfty}[{-p(nu)}(1+nu)(ln(1+nu))^2]\&=0end{align}$$ Provided $p(nu)$ is sufficiently rapidly decaying as $nutoinfty$ (which is a reasonable assumption to make, and assuming $|P(0)|<infty$, which is true since $p(nu)$ is a p.d.f, so the area under it between $0$ and $infty$ is finite.
$endgroup$
– John Doe
Dec 28 '18 at 5:04
|
show 2 more comments
$begingroup$
Integration by parts works as follows: $$int f'g=fg-int fg'$$.
That would explain the appearance of $$(log(1+nu))'=frac1{1+nu}$$.
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
add a comment |
$begingroup$
Integration by parts works as follows: $$int f'g=fg-int fg'$$.
That would explain the appearance of $$(log(1+nu))'=frac1{1+nu}$$.
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
add a comment |
$begingroup$
Integration by parts works as follows: $$int f'g=fg-int fg'$$.
That would explain the appearance of $$(log(1+nu))'=frac1{1+nu}$$.
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
$endgroup$
Integration by parts works as follows: $$int f'g=fg-int fg'$$.
That would explain the appearance of $$(log(1+nu))'=frac1{1+nu}$$.
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
edited Dec 28 '18 at 3:47
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
answered Dec 28 '18 at 3:36
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
add a comment |
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
$begingroup$
No. Notice there are two terms, one being an integral. The $fg$ term is zero, i guess.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:47
add a comment |
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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
$begingroup$
This is integration by parts.
$endgroup$
– Chris Custer
Dec 28 '18 at 3:21