Change of variable (translation) in complex integral
$begingroup$
If I have a real integral, e.g. $int f(x+2) dx$, I can substitute $y = x+2$, so $dy = dx$.
But if my function is complex, am I still allowed to do this? In which cases I cannot apply a translation?
integration complex-analysis multivariable-calculus
$endgroup$
add a comment |
$begingroup$
If I have a real integral, e.g. $int f(x+2) dx$, I can substitute $y = x+2$, so $dy = dx$.
But if my function is complex, am I still allowed to do this? In which cases I cannot apply a translation?
integration complex-analysis multivariable-calculus
$endgroup$
$begingroup$
Yes, it's still allowed. Whether $f$ is complex- or real-valued doesn't matter for this kind of transformations.
$endgroup$
– Daniel Fischer
Jan 18 '14 at 18:49
$begingroup$
@DanielFischer But I have to require $f$ holomorphic, right?
$endgroup$
– user119139
Jan 19 '14 at 15:53
$begingroup$
No, these transformations hold generally. You need to require $f$ holomorphic if you want to shift the contour or so, but not for reparametrisations like $$int_a^b f(x+2),dx = int_{a+2}^{b+2} f(y),dy.$$
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:04
$begingroup$
@DanielFischer But for example if $f$ has a pole in $0$, $$int_{-infty}^{+infty} f(x+i) dx neq int_{-infty}^{+infty} f(x-i) dx. $$
$endgroup$
– user119139
Jan 19 '14 at 16:44
1
$begingroup$
That is shifting the contour. $$int_{-infty}^{+infty} f(x+i),dx = int_{-infty+2i}^{+infty+2i} f(z-i),dz$$ would be the transformation that is valid without an assumption of holomorphy (but of course integrability; $f$ must be integrable on $(-infty+i,+infty+i)$).
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:50
add a comment |
$begingroup$
If I have a real integral, e.g. $int f(x+2) dx$, I can substitute $y = x+2$, so $dy = dx$.
But if my function is complex, am I still allowed to do this? In which cases I cannot apply a translation?
integration complex-analysis multivariable-calculus
$endgroup$
If I have a real integral, e.g. $int f(x+2) dx$, I can substitute $y = x+2$, so $dy = dx$.
But if my function is complex, am I still allowed to do this? In which cases I cannot apply a translation?
integration complex-analysis multivariable-calculus
integration complex-analysis multivariable-calculus
asked Jan 18 '14 at 18:43
user119139user119139
8217
8217
$begingroup$
Yes, it's still allowed. Whether $f$ is complex- or real-valued doesn't matter for this kind of transformations.
$endgroup$
– Daniel Fischer
Jan 18 '14 at 18:49
$begingroup$
@DanielFischer But I have to require $f$ holomorphic, right?
$endgroup$
– user119139
Jan 19 '14 at 15:53
$begingroup$
No, these transformations hold generally. You need to require $f$ holomorphic if you want to shift the contour or so, but not for reparametrisations like $$int_a^b f(x+2),dx = int_{a+2}^{b+2} f(y),dy.$$
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:04
$begingroup$
@DanielFischer But for example if $f$ has a pole in $0$, $$int_{-infty}^{+infty} f(x+i) dx neq int_{-infty}^{+infty} f(x-i) dx. $$
$endgroup$
– user119139
Jan 19 '14 at 16:44
1
$begingroup$
That is shifting the contour. $$int_{-infty}^{+infty} f(x+i),dx = int_{-infty+2i}^{+infty+2i} f(z-i),dz$$ would be the transformation that is valid without an assumption of holomorphy (but of course integrability; $f$ must be integrable on $(-infty+i,+infty+i)$).
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:50
add a comment |
$begingroup$
Yes, it's still allowed. Whether $f$ is complex- or real-valued doesn't matter for this kind of transformations.
$endgroup$
– Daniel Fischer
Jan 18 '14 at 18:49
$begingroup$
@DanielFischer But I have to require $f$ holomorphic, right?
$endgroup$
– user119139
Jan 19 '14 at 15:53
$begingroup$
No, these transformations hold generally. You need to require $f$ holomorphic if you want to shift the contour or so, but not for reparametrisations like $$int_a^b f(x+2),dx = int_{a+2}^{b+2} f(y),dy.$$
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:04
$begingroup$
@DanielFischer But for example if $f$ has a pole in $0$, $$int_{-infty}^{+infty} f(x+i) dx neq int_{-infty}^{+infty} f(x-i) dx. $$
$endgroup$
– user119139
Jan 19 '14 at 16:44
1
$begingroup$
That is shifting the contour. $$int_{-infty}^{+infty} f(x+i),dx = int_{-infty+2i}^{+infty+2i} f(z-i),dz$$ would be the transformation that is valid without an assumption of holomorphy (but of course integrability; $f$ must be integrable on $(-infty+i,+infty+i)$).
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:50
$begingroup$
Yes, it's still allowed. Whether $f$ is complex- or real-valued doesn't matter for this kind of transformations.
$endgroup$
– Daniel Fischer
Jan 18 '14 at 18:49
$begingroup$
Yes, it's still allowed. Whether $f$ is complex- or real-valued doesn't matter for this kind of transformations.
$endgroup$
– Daniel Fischer
Jan 18 '14 at 18:49
$begingroup$
@DanielFischer But I have to require $f$ holomorphic, right?
$endgroup$
– user119139
Jan 19 '14 at 15:53
$begingroup$
@DanielFischer But I have to require $f$ holomorphic, right?
$endgroup$
– user119139
Jan 19 '14 at 15:53
$begingroup$
No, these transformations hold generally. You need to require $f$ holomorphic if you want to shift the contour or so, but not for reparametrisations like $$int_a^b f(x+2),dx = int_{a+2}^{b+2} f(y),dy.$$
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:04
$begingroup$
No, these transformations hold generally. You need to require $f$ holomorphic if you want to shift the contour or so, but not for reparametrisations like $$int_a^b f(x+2),dx = int_{a+2}^{b+2} f(y),dy.$$
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:04
$begingroup$
@DanielFischer But for example if $f$ has a pole in $0$, $$int_{-infty}^{+infty} f(x+i) dx neq int_{-infty}^{+infty} f(x-i) dx. $$
$endgroup$
– user119139
Jan 19 '14 at 16:44
$begingroup$
@DanielFischer But for example if $f$ has a pole in $0$, $$int_{-infty}^{+infty} f(x+i) dx neq int_{-infty}^{+infty} f(x-i) dx. $$
$endgroup$
– user119139
Jan 19 '14 at 16:44
1
1
$begingroup$
That is shifting the contour. $$int_{-infty}^{+infty} f(x+i),dx = int_{-infty+2i}^{+infty+2i} f(z-i),dz$$ would be the transformation that is valid without an assumption of holomorphy (but of course integrability; $f$ must be integrable on $(-infty+i,+infty+i)$).
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:50
$begingroup$
That is shifting the contour. $$int_{-infty}^{+infty} f(x+i),dx = int_{-infty+2i}^{+infty+2i} f(z-i),dz$$ would be the transformation that is valid without an assumption of holomorphy (but of course integrability; $f$ must be integrable on $(-infty+i,+infty+i)$).
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:50
add a comment |
1 Answer
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$begingroup$
Daniel Fischer is correct. It is perfectly fine to do this (so long as you don't run into issues with poles). A more interesting question is evaluating real-line integrals that benefit from complex changes of variables. One thing people often do when taking the Fourier transform of the Gaussian, i.e. evaluating the integral (for $omegainmathbb{R}$)
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt,$$
is a complex change of variable and then equate it to a real-line integral. This implicitly relies on the function $exp(-z^2)$ being holomorphic. Here's how it's done traditionally:
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty} e^{frac{omega^2}{4}}e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty}e^{-left(-ifrac{omega}{2}-tright)^2}dt. $$
Then they say, let $y = -ifrac{omega}{2}-t$ and then state that this is equal to
$$int_{-infty}^{infty} e^{-y^2}dy$$
while wholly ignoring the complex nature of the line you are integrating over (you should be integrating over the line $y = -ifrac{omega}{2}$). But since the integrand is holomorphic, you can argue very simply that it is indeed equal to what they claim by parameterizing a rectangular contour with top and bottom being $y=-ifrac{omega}{2}$ and the real line, respectively (the sides can be shown to go to zero without too much difficulty). Since $exp(-z^2)$ is holomorphic, any closed contour of it will give $0$, hence the top and bottom pieces being equal and why they are correct to claim what they do.
If $exp(-z^2)$ were not holomorphic, we would need to invoke more involved complex analytic techniques to relate the complex line integral to the real line integral.
For completeness, we know that $int_{-infty}^{infty}e^{-y^2}dy = sqrt{pi}$ and so we see that
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = sqrt{pi}e^{-frac{omega^2}{4}}.$$
This is not the only way to evaluate this integral, but it is one of the two or three ways I've seen to argue that the Gaussian is what is called an eigenfunction of the Fourier transform (meaning the Fourier transform of the Gaussian gives a Gaussian with some scaling changes). Another way is by doing parametric differentiation on the initial integral but this requires some measure theoretic arguments.
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add a comment |
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$begingroup$
Daniel Fischer is correct. It is perfectly fine to do this (so long as you don't run into issues with poles). A more interesting question is evaluating real-line integrals that benefit from complex changes of variables. One thing people often do when taking the Fourier transform of the Gaussian, i.e. evaluating the integral (for $omegainmathbb{R}$)
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt,$$
is a complex change of variable and then equate it to a real-line integral. This implicitly relies on the function $exp(-z^2)$ being holomorphic. Here's how it's done traditionally:
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty} e^{frac{omega^2}{4}}e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty}e^{-left(-ifrac{omega}{2}-tright)^2}dt. $$
Then they say, let $y = -ifrac{omega}{2}-t$ and then state that this is equal to
$$int_{-infty}^{infty} e^{-y^2}dy$$
while wholly ignoring the complex nature of the line you are integrating over (you should be integrating over the line $y = -ifrac{omega}{2}$). But since the integrand is holomorphic, you can argue very simply that it is indeed equal to what they claim by parameterizing a rectangular contour with top and bottom being $y=-ifrac{omega}{2}$ and the real line, respectively (the sides can be shown to go to zero without too much difficulty). Since $exp(-z^2)$ is holomorphic, any closed contour of it will give $0$, hence the top and bottom pieces being equal and why they are correct to claim what they do.
If $exp(-z^2)$ were not holomorphic, we would need to invoke more involved complex analytic techniques to relate the complex line integral to the real line integral.
For completeness, we know that $int_{-infty}^{infty}e^{-y^2}dy = sqrt{pi}$ and so we see that
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = sqrt{pi}e^{-frac{omega^2}{4}}.$$
This is not the only way to evaluate this integral, but it is one of the two or three ways I've seen to argue that the Gaussian is what is called an eigenfunction of the Fourier transform (meaning the Fourier transform of the Gaussian gives a Gaussian with some scaling changes). Another way is by doing parametric differentiation on the initial integral but this requires some measure theoretic arguments.
$endgroup$
add a comment |
$begingroup$
Daniel Fischer is correct. It is perfectly fine to do this (so long as you don't run into issues with poles). A more interesting question is evaluating real-line integrals that benefit from complex changes of variables. One thing people often do when taking the Fourier transform of the Gaussian, i.e. evaluating the integral (for $omegainmathbb{R}$)
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt,$$
is a complex change of variable and then equate it to a real-line integral. This implicitly relies on the function $exp(-z^2)$ being holomorphic. Here's how it's done traditionally:
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty} e^{frac{omega^2}{4}}e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty}e^{-left(-ifrac{omega}{2}-tright)^2}dt. $$
Then they say, let $y = -ifrac{omega}{2}-t$ and then state that this is equal to
$$int_{-infty}^{infty} e^{-y^2}dy$$
while wholly ignoring the complex nature of the line you are integrating over (you should be integrating over the line $y = -ifrac{omega}{2}$). But since the integrand is holomorphic, you can argue very simply that it is indeed equal to what they claim by parameterizing a rectangular contour with top and bottom being $y=-ifrac{omega}{2}$ and the real line, respectively (the sides can be shown to go to zero without too much difficulty). Since $exp(-z^2)$ is holomorphic, any closed contour of it will give $0$, hence the top and bottom pieces being equal and why they are correct to claim what they do.
If $exp(-z^2)$ were not holomorphic, we would need to invoke more involved complex analytic techniques to relate the complex line integral to the real line integral.
For completeness, we know that $int_{-infty}^{infty}e^{-y^2}dy = sqrt{pi}$ and so we see that
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = sqrt{pi}e^{-frac{omega^2}{4}}.$$
This is not the only way to evaluate this integral, but it is one of the two or three ways I've seen to argue that the Gaussian is what is called an eigenfunction of the Fourier transform (meaning the Fourier transform of the Gaussian gives a Gaussian with some scaling changes). Another way is by doing parametric differentiation on the initial integral but this requires some measure theoretic arguments.
$endgroup$
add a comment |
$begingroup$
Daniel Fischer is correct. It is perfectly fine to do this (so long as you don't run into issues with poles). A more interesting question is evaluating real-line integrals that benefit from complex changes of variables. One thing people often do when taking the Fourier transform of the Gaussian, i.e. evaluating the integral (for $omegainmathbb{R}$)
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt,$$
is a complex change of variable and then equate it to a real-line integral. This implicitly relies on the function $exp(-z^2)$ being holomorphic. Here's how it's done traditionally:
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty} e^{frac{omega^2}{4}}e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty}e^{-left(-ifrac{omega}{2}-tright)^2}dt. $$
Then they say, let $y = -ifrac{omega}{2}-t$ and then state that this is equal to
$$int_{-infty}^{infty} e^{-y^2}dy$$
while wholly ignoring the complex nature of the line you are integrating over (you should be integrating over the line $y = -ifrac{omega}{2}$). But since the integrand is holomorphic, you can argue very simply that it is indeed equal to what they claim by parameterizing a rectangular contour with top and bottom being $y=-ifrac{omega}{2}$ and the real line, respectively (the sides can be shown to go to zero without too much difficulty). Since $exp(-z^2)$ is holomorphic, any closed contour of it will give $0$, hence the top and bottom pieces being equal and why they are correct to claim what they do.
If $exp(-z^2)$ were not holomorphic, we would need to invoke more involved complex analytic techniques to relate the complex line integral to the real line integral.
For completeness, we know that $int_{-infty}^{infty}e^{-y^2}dy = sqrt{pi}$ and so we see that
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = sqrt{pi}e^{-frac{omega^2}{4}}.$$
This is not the only way to evaluate this integral, but it is one of the two or three ways I've seen to argue that the Gaussian is what is called an eigenfunction of the Fourier transform (meaning the Fourier transform of the Gaussian gives a Gaussian with some scaling changes). Another way is by doing parametric differentiation on the initial integral but this requires some measure theoretic arguments.
$endgroup$
Daniel Fischer is correct. It is perfectly fine to do this (so long as you don't run into issues with poles). A more interesting question is evaluating real-line integrals that benefit from complex changes of variables. One thing people often do when taking the Fourier transform of the Gaussian, i.e. evaluating the integral (for $omegainmathbb{R}$)
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt,$$
is a complex change of variable and then equate it to a real-line integral. This implicitly relies on the function $exp(-z^2)$ being holomorphic. Here's how it's done traditionally:
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty} e^{frac{omega^2}{4}}e^{-iomega t}e^{-t^2}dt = e^{-frac{omega^2}{4}}int_{-infty}^{infty}e^{-left(-ifrac{omega}{2}-tright)^2}dt. $$
Then they say, let $y = -ifrac{omega}{2}-t$ and then state that this is equal to
$$int_{-infty}^{infty} e^{-y^2}dy$$
while wholly ignoring the complex nature of the line you are integrating over (you should be integrating over the line $y = -ifrac{omega}{2}$). But since the integrand is holomorphic, you can argue very simply that it is indeed equal to what they claim by parameterizing a rectangular contour with top and bottom being $y=-ifrac{omega}{2}$ and the real line, respectively (the sides can be shown to go to zero without too much difficulty). Since $exp(-z^2)$ is holomorphic, any closed contour of it will give $0$, hence the top and bottom pieces being equal and why they are correct to claim what they do.
If $exp(-z^2)$ were not holomorphic, we would need to invoke more involved complex analytic techniques to relate the complex line integral to the real line integral.
For completeness, we know that $int_{-infty}^{infty}e^{-y^2}dy = sqrt{pi}$ and so we see that
$$int_{-infty}^{infty} e^{-iomega t}e^{-t^2}dt = sqrt{pi}e^{-frac{omega^2}{4}}.$$
This is not the only way to evaluate this integral, but it is one of the two or three ways I've seen to argue that the Gaussian is what is called an eigenfunction of the Fourier transform (meaning the Fourier transform of the Gaussian gives a Gaussian with some scaling changes). Another way is by doing parametric differentiation on the initial integral but this requires some measure theoretic arguments.
edited Jan 9 at 7:11
answered Jan 18 '14 at 19:05
Cameron WilliamsCameron Williams
22.6k43680
22.6k43680
add a comment |
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$begingroup$
Yes, it's still allowed. Whether $f$ is complex- or real-valued doesn't matter for this kind of transformations.
$endgroup$
– Daniel Fischer
Jan 18 '14 at 18:49
$begingroup$
@DanielFischer But I have to require $f$ holomorphic, right?
$endgroup$
– user119139
Jan 19 '14 at 15:53
$begingroup$
No, these transformations hold generally. You need to require $f$ holomorphic if you want to shift the contour or so, but not for reparametrisations like $$int_a^b f(x+2),dx = int_{a+2}^{b+2} f(y),dy.$$
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:04
$begingroup$
@DanielFischer But for example if $f$ has a pole in $0$, $$int_{-infty}^{+infty} f(x+i) dx neq int_{-infty}^{+infty} f(x-i) dx. $$
$endgroup$
– user119139
Jan 19 '14 at 16:44
1
$begingroup$
That is shifting the contour. $$int_{-infty}^{+infty} f(x+i),dx = int_{-infty+2i}^{+infty+2i} f(z-i),dz$$ would be the transformation that is valid without an assumption of holomorphy (but of course integrability; $f$ must be integrable on $(-infty+i,+infty+i)$).
$endgroup$
– Daniel Fischer
Jan 19 '14 at 16:50