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Sign convention for Fourier transform and contour integration - example

1

$begingroup$

I was wondering about one (probably trivial) fact during computing the Fourier transform while using contour integral.

As an example I have following function:

$$f(x)={{1}over{x^2+a^2}}$$ and its Fourier transform is equal to: $$FT[f(x)](s)={sqrt{piover 2}}e^{-|s|a}.$$

I used this definition of Fourier transform:

$$ FT[f(x)](s) =frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(x){e^{-isx}dx}$$

The singularities are $ia$ and $-ia$ and we have the solutions for $s lt 0$ and $s gt 0$. And we are applying the Jordan's lemma.

But why for $s lt 0$ we integrate along upper half-plane and for $s gt 0$ we have the lower half-plane? I confuse the sign notation. Everytime I think I have figured it out I have feeling it is not a real explanation, even though the reason is probably pretty simple. But every textbook or internet source just says it is like that and not why.

calculus complex-analysis fourier-analysis contour-integration fourier-transform

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asked Mar 22 at 13:18

LeifLeif

628314

$endgroup$

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1

$begingroup$

I was wondering about one (probably trivial) fact during computing the Fourier transform while using contour integral.

As an example I have following function:

$$f(x)={{1}over{x^2+a^2}}$$ and its Fourier transform is equal to: $$FT[f(x)](s)={sqrt{piover 2}}e^{-|s|a}.$$

I used this definition of Fourier transform:

$$ FT[f(x)](s) =frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(x){e^{-isx}dx}$$

The singularities are $ia$ and $-ia$ and we have the solutions for $s lt 0$ and $s gt 0$. And we are applying the Jordan's lemma.

But why for $s lt 0$ we integrate along upper half-plane and for $s gt 0$ we have the lower half-plane? I confuse the sign notation. Everytime I think I have figured it out I have feeling it is not a real explanation, even though the reason is probably pretty simple. But every textbook or internet source just says it is like that and not why.

calculus complex-analysis fourier-analysis contour-integration fourier-transform

share|cite|improve this question

asked Mar 22 at 13:18

LeifLeif

628314

$endgroup$

add a comment | 

$begingroup$

I was wondering about one (probably trivial) fact during computing the Fourier transform while using contour integral.

As an example I have following function:

$$f(x)={{1}over{x^2+a^2}}$$ and its Fourier transform is equal to: $$FT[f(x)](s)={sqrt{piover 2}}e^{-|s|a}.$$

I used this definition of Fourier transform:

$$ FT[f(x)](s) =frac{1}{sqrt{2 pi}}int_{-infty}^{infty}f(x){e^{-isx}dx}$$

The singularities are $ia$ and $-ia$ and we have the solutions for $s lt 0$ and $s gt 0$. And we are applying the Jordan's lemma.

But why for $s lt 0$ we integrate along upper half-plane and for $s gt 0$ we have the lower half-plane? I confuse the sign notation. Everytime I think I have figured it out I have feeling it is not a real explanation, even though the reason is probably pretty simple. But every textbook or internet source just says it is like that and not why.

calculus complex-analysis fourier-analysis contour-integration fourier-transform

share|cite|improve this question

asked Mar 22 at 13:18

LeifLeif

628314

$endgroup$

share|cite|improve this question

asked Mar 22 at 13:18

LeifLeif

628314

share|cite|improve this question

asked Mar 22 at 13:18

LeifLeif

628314

asked Mar 22 at 13:18

LeifLeif

628314

asked Mar 22 at 13:18

LeifLeif

628314

1 Answer
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2

$begingroup$

Let $x=a+bi$ Then we have that
$$exp(-ixs)=exp(-i(a+bi)s)=exp(-ias+bs)=exp(-ias)exp(bs)$$
If $s in mathbb{R}$, then the first term is oscillating. If $s>0$, then $b$ should be negative (otherwise $exp(bs) to infty$) to have a finite integral. If $s<0$, we should have positive $b$. And $b$ is positive in the upper half plane, and negative in the bottom half plane.

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answered Mar 22 at 13:29

BotondBotond

6,57531034

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2

$begingroup$

Let $x=a+bi$ Then we have that
$$exp(-ixs)=exp(-i(a+bi)s)=exp(-ias+bs)=exp(-ias)exp(bs)$$
If $s in mathbb{R}$, then the first term is oscillating. If $s>0$, then $b$ should be negative (otherwise $exp(bs) to infty$) to have a finite integral. If $s<0$, we should have positive $b$. And $b$ is positive in the upper half plane, and negative in the bottom half plane.

share|cite|improve this answer

answered Mar 22 at 13:29

BotondBotond

6,57531034

$endgroup$

add a comment | 

2

$begingroup$

Let $x=a+bi$ Then we have that
$$exp(-ixs)=exp(-i(a+bi)s)=exp(-ias+bs)=exp(-ias)exp(bs)$$
If $s in mathbb{R}$, then the first term is oscillating. If $s>0$, then $b$ should be negative (otherwise $exp(bs) to infty$) to have a finite integral. If $s<0$, we should have positive $b$. And $b$ is positive in the upper half plane, and negative in the bottom half plane.

share|cite|improve this answer

answered Mar 22 at 13:29

BotondBotond

6,57531034

$endgroup$

add a comment | 

$begingroup$

Let $x=a+bi$ Then we have that
$$exp(-ixs)=exp(-i(a+bi)s)=exp(-ias+bs)=exp(-ias)exp(bs)$$
If $s in mathbb{R}$, then the first term is oscillating. If $s>0$, then $b$ should be negative (otherwise $exp(bs) to infty$) to have a finite integral. If $s<0$, we should have positive $b$. And $b$ is positive in the upper half plane, and negative in the bottom half plane.

share|cite|improve this answer

answered Mar 22 at 13:29

BotondBotond

6,57531034

$endgroup$

share|cite|improve this answer

answered Mar 22 at 13:29

BotondBotond

6,57531034

answered Mar 22 at 13:29

BotondBotond

6,57531034

answered Mar 22 at 13:29

BotondBotond

6,57531034