Help solving this question (Fourier Transform)

[YehiaMedhat]

The question is to solve the following integral: [imath]\int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} dx[/imath]
I tried to think of it like convolution, but this needs one more integral like this [imath]\int_{-\infty}^{\infty}[/imath], so this doesn't look like a good one.
I tried to think of main formula of the Fourier transform, which is [imath]\int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt[/imath] and letting [imath]\omega[/imath] be 0, so it would end up like [imath]\int_{-\infty}^{\infty} f(t) dt[/imath], but I still have two functions, and the first looks like differentiated. I mean how could I get the Fourier transform for [imath]e^{-3|t-x|}[/imath] and [imath]\frac{x}{(x^2 + 4)^2}[/imath] together without seeing the conditions for convolution.

If it's, potentially, convolution it should have looked like this: [imath]\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} dx[/imath], shouldn't it?

Any one who have hints, or who could nudge me in the right direction to solving this problem, or just referring some text book which has similar examples, that will be a great help for me.

 

[blamocur]

Hint : break the original integral into two to get rid of absolute value in the exponent.

 

[YehiaMedhat]

Hint : break the original integral into two to get rid of absolute value in the exponent.

Ok, trying to do that, I get the following:
[math]\int_{-\infty}^{t} \frac{x}{(x^2 + 4)^2} e^{3(x-t)} dx + \int_{t}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3(x-t)} dx[/math]Here, I think that the first integral is using the time integration property, but still, where is the main integral of the Fourier transform itself? so that looks obselete.

Thinking about integration by parts, both functions don't show some potential to be solved by recursion, so that also seems like not a good idea.

I'm stuck in it, I don't know. Please one more hint

 

[YehiaMedhat]

Ok, I got someone else telling me to get the Fourier transform of it and inverse it after simplification, but doing so, it seemed to get harder. Here is what I've got:
$\mathcal{F}\left\{ \int_{-\infty}^{\infty} \frac{x}{\left(x^2+4\right)^2} e^{-3|t-x|} dx\right\} = \mathcal{F}\left\{ \frac{x}{\left(x^2+4\right)^2} \right\} \mathcal{F}\left\{ e^{-3|x|} \right\}$$

Which leads to:
[math]\left(\frac{6}{9 + \omega^2}\right) \frac{-1}{2}\mathcal{F}\left\{\frac{-2x}{(\omega^2+4)^2}\right\}[/math][math]= \left(\frac{6}{9 + \omega^2}\right) \frac{-1}{2i}\mathcal{F}\left\{\frac{d}{d\omega}\frac{1}{\omega^2+4}\right\}[/math][math]= \left(\frac{6}{9 + \omega^2}\right) \frac{i}{8}\mathcal{F}\left\{\frac{d}{d\omega}\frac{4}{\omega^2+4}\right\}[/math][math]= \frac{i}{8} \left(\frac{6}{9 + \omega^2}\right) \left( -2\pi \omega e^{-2|\omega|}\right)[/math]
, isn't it? just I need to get the inverse of the Fourier transform now, so trying as follows:
[math]\frac{-3i\pi}{4} \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{9 + \omega^2}\ \omega e^{-2|\omega|}\ e^{i\omega t} d\omega[/math][math]\frac{-3i\pi}{4} \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{9 + \omega^2}\ \omega e^{-2|\omega|+ i\omega t}\ d\omega[/math][math]\frac{-3i}{8} \left[\int_{-\infty}^{0} \frac{1}{9 + \omega^2}\ \omega e^{(2+ it)\omega}\ d\omega + \int_{0}^{\infty} \frac{1}{9 + \omega^2}\ \omega e^{(-2+ it)\omega}\ d\omega \right][/math]
After this, how should I proceed? since I don't feel like this integral is doable with my knowledge of integration by parts or even using some recursive formula.

 

[blamocur]

Thinking about integration by parts, both functions don't show some potential to be solved by recursion

I've noticed that [imath]\frac{x dx}{(x^2+4)^2}[/imath] is easy to integrate, so integration by parts transforms the whole thing to something slightly simpler, but still not simple enough for me to solve Nor do I remember enough of Fourier transform properties to know how to apply it to you problem.

Just curious: is this from homework or something else?

 

[mario99]

Ok, I got someone else telling me to get the Fourier transform of it and inverse it after simplification, but doing so, it seemed to get harder. Here is what I've got:
$\mathcal{F}\left\{ \int_{-\infty}^{\infty} \frac{x}{\left(x^2+4\right)^2} e^{-3|t-x|} dx\right\} = \mathcal{F}\left\{ \frac{x}{\left(x^2+4\right)^2} \right\} \mathcal{F}\left\{ e^{-3|x|} \right\}$$

Which leads to:
[math]\left(\frac{6}{9 + \omega^2}\right) \frac{-1}{2}\mathcal{F}\left\{\frac{-2x}{(\omega^2+4)^2}\right\}[/math][math]= \left(\frac{6}{9 + \omega^2}\right) \frac{-1}{2i}\mathcal{F}\left\{\frac{d}{d\omega}\frac{1}{\omega^2+4}\right\}[/math][math]= \left(\frac{6}{9 + \omega^2}\right) \frac{i}{8}\mathcal{F}\left\{\frac{d}{d\omega}\frac{4}{\omega^2+4}\right\}[/math][math]= \frac{i}{8} \left(\frac{6}{9 + \omega^2}\right) \left( -2\pi \omega e^{-2|\omega|}\right)[/math]
, isn't it? just I need to get the inverse of the Fourier transform now, so trying as follows:
[math]\frac{-3i\pi}{4} \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{9 + \omega^2}\ \omega e^{-2|\omega|}\ e^{i\omega t} d\omega[/math][math]\frac{-3i\pi}{4} \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{9 + \omega^2}\ \omega e^{-2|\omega|+ i\omega t}\ d\omega[/math][math]\frac{-3i}{8} \left[\int_{-\infty}^{0} \frac{1}{9 + \omega^2}\ \omega e^{(2+ it)\omega}\ d\omega + \int_{0}^{\infty} \frac{1}{9 + \omega^2}\ \omega e^{(-2+ it)\omega}\ d\omega \right][/math]
After this, how should I proceed? since I don't feel like this integral is doable with my knowledge of integration by parts or even using some recursive formula.

This does not make sense because the idea of convolution is that you have an inverse Fourier transform that you cannot solve, then you change it to convolution which is easier to handle. Not the reverse!

[imath]\displaystyle f(x) = \mathcal{F^{-1}}\{F(\omega)\} = \int_{-\infty}^{\infty} F(\omega)e^{-i\omega x} \ d\omega[/imath]

The idea is that we don't have the inverse Fourier transform of [imath]F(\omega)[/imath], then we let [imath]F(\omega) = U(\omega)V(\omega)[/imath]. We have (know) the inverse Fourier transform of [imath]U(\omega)[/imath] and the inverse Fourier transform of [imath]V(\omega)[/imath].

[imath]\mathcal{F^{-1}}\{U(\omega)\} = u(x)[/imath]
[imath]\mathcal{F^{-1}}\{V(\omega)\} = v(x)[/imath]

Now we can change the inverse Fourier transform to convolution:

[imath]\displaystyle f(x) = \mathcal{F^{-1}}\{F(\omega)\} = \mathcal{F^{-1}}\{U(\omega)V(\omega)\} = u(x)*v(x) = \int_{-\infty}^{\infty} u(\tau)v(x-\tau) \ d\tau[/imath]

 

[YehiaMedhat]

Hey guys, please I'm screwed and have been asking and following forms and responses, but only gets worse.
Please can any one tell me what should be the way.

I have tried to do the Fourier transform of the convolution form as it's kinda counter intuitive to do so, and then apply symmetry for the first fraction and get the transform of the exponential but it just got more complicated.

I was doing research all the day!

 

[YehiaMedhat]

I've noticed that [imath]\frac{x dx}{(x^2+4)^2}[/imath] is easy to integrate, so integration by parts transforms the whole thing to something slightly simpler, but still not simple enough for me to solve Nor do I remember enough of Fourier transform properties to know how to apply it to you problem.

Just curious: is this from homework or something else?

Yes it's a sheet, I'm an electrical power engineering student btw

 

[mario99]

Hey guys, please I'm screwed and have been asking and following forms and responses, but only gets worse.
Please can any one tell me what should be the way.

I have tried to do the Fourier transform of the convolution form as it's kinda counter intuitive to do so, and then apply symmetry for the first fraction and get the transform of the exponential but it just got more complicated.

I was doing research all the day!

I did a contour integration and I got this.

[imath]\displaystyle \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} \ dx = \frac{3\pi}{2}e^{-i6}\sinh 3t[/imath]

This is my result assuming that my calculation went right all the way!

 

[blamocur]

I did a contour integration and I got this.

[imath]\displaystyle \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} \ dx = \frac{3\pi}{2}e^{-i6}\sinh 3t[/imath]

This is my result assuming that my calculation went right all the way!

Your right hand side is not real, but the left hand side should be, at least for real [imath]t[/imath]'s.

 

[mario99]

Your right hand side is not real, but the left hand side should be, at least for real [imath]t[/imath]'s.

I have included the real and imaginary parts in purpose. I wanna the OP to compare my solution with the solution of the sheet and to take whatever he needs. Let us just hope that he has the final solution!

 

[blamocur]

I have included the real and imaginary parts in purpose. I wanna the OP to compare my solution with the solution of the sheet and to take whatever he needs. Let us just hope that he has the final solution!

But how can you have an equality between real and non-real values?

 

[mario99]

But how can you have an equality between real and non-real values?

Some problems asks for both parts of the contour. Have you ever solved an integration by contours?

I understand what you mean. The original integral is real, so it is wrong to write it as [imath] a + bi[/imath].

I should have written as:

[imath]\displaystyle \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} \ dx = \frac{3\pi}{2}\cos 6\sinh 3t[/imath]

 

[YehiaMedhat]

Some problems asks for both parts of the contour. Have you ever solved an integration by contours?

I understand what you mean. The original integral is real, so it is wrong to write it as [imath] a + bi[/imath].

Since the original integral is real, I should have written:

[imath]\displaystyle \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} \ dx = \frac{3\pi}{2}\cos 6\sinh 3t[/imath]

Well so it seems like I haven't taken this kind of integration yet. Overall, thank you all guys

 

[mario99]

Well so it seems like I haven't taken this kind of integration yet. Overall, thank you all guys

I am wondering from where did you get that convolution integral! How did they ask you to solve an integral that was beyond your capabilities? Do you have the solution to that problem?

It is awkward. Are we missing something?

 

[blamocur]

Some problems asks for both parts of the contour. Have you ever solved an integration by contours?

I understand what you mean. The original integral is real, so it is wrong to write it as [imath] a + bi[/imath].

I should have written as:

[imath]\displaystyle \int_{-\infty}^{\infty} \frac{x}{(x^2 + 4)^2} e^{-3|t-x|} \ dx = \frac{3\pi}{2}\cos 6\sinh 3t[/imath]

No, you shouldn't

More seriously: one can show that the left-hand side is limited (by about 1/30 actually), but the right-hand side is proportional to [imath]\sinh 3t[/imath], which can grow infinitely for large [imath]t[/imath]'s.

 

[mario99]

No, you shouldn't

More seriously: one can show that the left-hand side is limited (by about 1/30 actually), but the right-hand side is proportional to [imath]\sinh 3t[/imath], which can grow infinitely for large [imath]t[/imath]'s.

Professor blamocur. Now I am confused. You don't want me to write the solution with the imaginary part but in the same time you say I should not. What do you mean exactly? Please explain!

 

[blamocur]

Professor blamocur. Now I am confused. You don't want me to write the solution with the imaginary part but in the same time you say I should not. What do you mean exactly? Please explain!

And I am confused too: what kind of explanation do you need? I don't "want" you to write anything, if only because I don't know the answer myself. But if you claim that the two expressions are equal for any [imath]t[/imath], and I show you that they cannot be equal, is not that enough for you to reconsider your posts?

 

[mario99]

And I am confused too: what kind of explanation do you need? I don't "want" you to write anything, if only because I don't know the answer myself. But if you claim that the two expressions are equal for any [imath]t[/imath], and I show you that they cannot be equal, is not that enough for you to reconsider your posts?

If you know the correct answer of the problem, why not sharing it, so we could see where I went wrong. The OP has already shared enough tries. And I did my solution based on contour integration. I asked the OP if he has the solution (to compare), but he did not answer that. And I asked you too (to compare), have you ever solved contour integration, but you did not answer that too.

Where is the problem in sharing what you know? If it is wrong, we can correct it. If it is almost correct, we can improve it. You don't have to be 100% correct to speak.

In fact, you did not show me they cannot be equal. You only explained it in a way that you think others would understand what you were saying. I understand the solution only when you write it down in full steps by your favorite method, not only by explaining it due to your high knowledge in integrals where you can see the answer can make sense or not.

I hope that I have not misunderstood your reply.

 

[blamocur]

And I asked you too (to compare), have you ever solved contour integration, but you did not answer that too.

I've mentioned more than once that I don't know the answer. E.g., from my post #18: "if only because I don't know the answer myself." Verifying solutions is often much easier than finding them - look up NP problems in computer science, for example.

In fact, you did not show me they cannot be equal.

I did my best, but please let me know which part you did not find convincing and why -- we can then discuss it in more detail.

I hope that I have not misunderstood your reply.

Sorry, but it looks to me like you have.