Inverse Fourier Transform for sine and cosine

[wolly]

Hi I want to find the inverse of the sine and cosine functions in the Fourier Transform

$$f(x)=\frac{\sqrt{2}}{\sqrt{\pi}} \int_{0}^{\infty} F_{C}(ω)cos(ωt)dω$$
and

$$f(x)=\frac{\sqrt{2}}{\sqrt{\pi}} \int_{0}^{\infty} F_{C}(ω)sin(ωt)dω$$
How can these 2 get this result?

 

[logistic_guy]

Hi I want to find the inverse of the sine and cosine functions in the Fourier Transform

I think that you mean here that you want to find the inverse of a function using the sine and cosine Fourier Transforms.

How can these 2 get this result?

What do you mean?

In the picture, they want to find the inverse Fourier transform of $\displaystyle \frac{1}{1 + t^2}$ using the cosine Fourier Transforms.

 

[wolly]

I think that you mean here that you want to find the inverse of a function using the sine and cosine Fourier Transforms.


What do you mean?

In the picture, they want to find the inverse Fourier transform of $\displaystyle \frac{1}{1 + t^2}$ using the cosine Fourier Transforms.

What I mean îs how did $$\frac{1}{\pi}$$ and $$\frac{2}{\pi}$$ appear in that integral ?when the inverse of Fourier functions are posted above.

 

[wolly]

I think that you mean here that you want to find the inverse of a function using the sine and cosine Fourier Transform

Exactly.Is there a method?

 

[wolly]

In the picture, they want to find the inverse Fourier transform of $\displaystyle \frac{1}{1 + t^2}$ using the cosine Fourier Transforms.

And another thing
How did This function transform into

$$g(z)= \frac{e^{izx}}{z^2+1}$$?

 

[wolly]

In my book This exercise îs solved by Jordan lemma and I Don't know how to apply it

 

[wolly]

Is this based on the Fourier Integral?
@logistic_guy ??

 

[logistic_guy]

What I mean îs how did $$\frac{1}{\pi}$$ and $$\frac{2}{\pi}$$ appear in that integral ?when the inverse of Fourier functions are posted above.

$\displaystyle \frac{2}{\pi}$ is part of the inverse sine or cosine Fourier transform.

If you have $\displaystyle \int_{-\infty}^{\infty}$, you can take half the domain and double the integral, so $\displaystyle \int_{-\infty}^{\infty} = 2\int_{0}^{\infty}\$ Or $\displaystyle \ 2\int_{0}^{\infty} = \int_{-\infty}^{\infty}$


This is why $\displaystyle \frac{2}{\pi}\int_{0}^{\infty} = \frac{1}{\pi}\int_{-\infty}^{\infty}$

And another thing
How did This function transform into

$$g(z)= \frac{e^{izx}}{z^2+1}$$?

If the domain is $\displaystyle -\infty < z < \infty$, use the general Fourier transform:

$\displaystyle f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(z) e^{izx} \ dz$


If you want to work in the domain is $\displaystyle 0 < z < \infty$, and $\displaystyle f(z)$ is an even function, you can use the cosine Fourier transform:

$\displaystyle f(x) = \frac{2}{\pi}\int_{0}^{\infty} f(z) \cos(zx) \ dz = \frac{1}{\pi}\int_{-\infty}^{\infty} f(z) \cos(zx) \ dz$


If you want to work in the domain is $\displaystyle 0 < z < \infty$, and $\displaystyle f(z)$ is an odd function, you can use the sine Fourier transform:

$\displaystyle f(x) = \frac{2}{\pi}\int_{0}^{\infty} f(z) \sin(zx) \ dz = \frac{1}{\pi}\int_{-\infty}^{\infty} f(z) \sin(zx) \ dz$


Now let us apply the idea on $\displaystyle f(z) = \frac{1}{z^2 + 1}$. To find the inverse Fourier transform of this function, we usually use the general Fourier transform.

$\displaystyle f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{z^2 + 1} e^{izx} \ dz$

We know that $\displaystyle f(z) = \frac{1}{z^2 + 1}$ is an even function, so we have the option to solve it by the cosine Fourier transform on the domain $\displaystyle 0 < z < \infty$.

$\displaystyle f(x) = \frac{2}{\pi}\int_{0}^{\infty} \frac{1}{z^2 + 1} \cos(zx) \ dz = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{z^2 + 1} \cos(zx) \ dz = e^{-x}$

But this answer is only for $\displaystyle x > 0$. What about $\displaystyle x<0$? Therefore the full answer of the transform on $\displaystyle -\infty < z < \infty$ and $\displaystyle -\infty < x < \infty$ is:

$\displaystyle \mathcal{F^{-1}}\left\{\frac{1}{z^2 + 1}\right\} = \begin{cases}e^{-x} & x > 0\\e^{x} & x < 0\end{cases} = e^{-|x|}$

 

[logistic_guy]

In my book This exercise îs solved by Jordan lemma and I Don't know how to apply it

I will explain to you the idea of the Jordan lemma. If you solve integrals in complex analysis, you use a contour (path). This contour splits the original integral into two or more integrals. One of these integrals involves an arc (usually an arc of a semicircle). The Jordan lemma says that the integral that involves the arc $\displaystyle = 0$. That's the whole idea.

Now let us apply it step by step.

$$g(z)= \frac{e^{izx}}{z^2+1}$$?

First step is that we have to define our complex function.

Let our complex function be $\displaystyle g(z)= \frac{e^{izx}}{z^2+1}$.

We realize that this function has two poles $\displaystyle z = i$ and $\displaystyle z = -i$

Step 2. We will choose a contour. Let our contour be $\displaystyle \gamma$. We will let this contour contains only one of the poles. What about the top part of a semicircle of radius $\displaystyle R$ and $\displaystyle R > 1$. This means that the pole $\displaystyle z = i$ will be inside the contour (the semicircle) and we can compute its residue.

This is our contour: We will walk from $\displaystyle z = -R$ to $\displaystyle z = R$ (real line), then from $\displaystyle z = R$ back to $\displaystyle z = -R$ through the semicircular arc which we will call $\displaystyle \gamma_R$. As you can see, our contour is perfectly closed.

Then, our integral is:

$\displaystyle \int_{\gamma} g(z) \ dz = \int_{-R}^{R} g(z) \ dz + \int_{\gamma_R} g(z) \ dz$


Step 3. Apply the residue theorem.

$\displaystyle \int_{\gamma} g(z) \ dz = 2\pi i \cdot \text{Res}(g,i)$

This means that:

$\displaystyle \int_{-R}^{R} g(z) \ dz + \int_{\gamma_R} g(z) \ dz = 2\pi i \cdot \text{Res}(g,i)$


Step 4. Apply the limit to both sides.

$\displaystyle \lim_{R \rightarrow \infty}\int_{-R}^{R} g(z) \ dz + \lim_{R \rightarrow \infty} \int_{\gamma_R} g(z) \ dz = \lim_{R \rightarrow \infty} 2\pi i \cdot \text{Res}(g,i)$


Step 5. Use the Jordan lemma.

$\displaystyle \lim_{R \rightarrow \infty} \int_{\gamma_R} g(z) \ dz = 0$

Then,

$\displaystyle \lim_{R \rightarrow \infty}\int_{-R}^{R} g(z) \ dz = \lim_{R \rightarrow \infty} 2\pi i \cdot \text{Res}(g,i)$

Or

$\displaystyle \lim_{R \rightarrow \infty}\int_{-R}^{R} g(z) \ dz = 2\pi i \cdot \text{Res}(g,i)$


Step 6. Calculate the residue.

$\displaystyle \text{Res}(g,i) = \frac{e^{-x}}{2i}$

Then,

$\displaystyle \lim_{R \rightarrow \infty}\int_{-R}^{R} g(z) \ dz = \pi e^{-x}$

Or

$\displaystyle \int_{\gamma} g(z) \ dz = \lim_{R \rightarrow \infty}\int_{-R}^{R} g(z) \ dz = \pi e^{-x}$

Or

$\displaystyle \int_{-\infty}^{\infty} \frac{e^{izx}}{z^2 + 1} \ dz = \pi e^{-x}$

Since we have chosen the contour as the upper half of the semicircle, we were actually calculating only $\displaystyle x > 0$.

In other words, we were calculating the cosine Fourier transform.

Then,

$\displaystyle \int_{-\infty}^{\infty}\frac{\cos(xz)}{z^2 + 1} \ dz = \text{Re}\left(\int_{-\infty}^{\infty} \frac{e^{izx}}{z^2 + 1} \ dz\right) = \pi e^{-x}$

Or

$\displaystyle \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\cos(xz)}{z^2 + 1} \ dz = e^{-x}$

The same result we have obtained in post #$\displaystyle 8$.

Now if you wanna calculate $\displaystyle x < 0$ using the Jordan lemma, you do the same calculations we have done but you choose your contour to be the bottom part of the semicircle.

You will get: $\displaystyle \int_{-\infty}^{\infty} \frac{e^{izx}}{z^2 + 1} \ dz = \pi e^{x}$

And by combining the two solutions together, you will get again:

$\displaystyle f(x) = \mathcal{F^{-1}}\left\{\frac{1}{z^2 + 1}\right\} =\begin{cases}e^{-x} & x > 0\\e^{x} & x < 0\end{cases} = e^{-|x|}$