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I am trying to calculate integrals of the form:

$$ I(t, n) = \int_{-\infty}^{\infty} \Big(\frac{1}{1-jq}\big)^{n} e^{-jqt} dq $$

where $j = \sqrt{-1}$. In the case when $n=1$, I have:

$$ I(t, 1) = \int_{-\infty}^{\infty} \frac{1}{1-jq} e^{-jqt} dq $$

Now, my thought was to view this as a function of $t$, and use the Feynmann trick. Ie:

$$ \frac{d}{dt} I(t, 1) = \int_{-\infty}^{\infty} \frac{\partial}{\partial t} \big(\frac{1}{1-jq} e^{-jqt} \big) dq $$ $$ \frac{d}{dt} I(t, 1) = \int_{-\infty}^{\infty} \frac{-jq}{1-jq} e^{-jqt} dq $$

This looks promising, but I can't get it to go anywhere. An even more promising avenue seems to be to express the exponential as a power series around zero. This gives:

$$ I(t, 1) = \int_{-\infty}^{\infty} \frac{1}{1-jq} \sum_{k=0}^{\infty} \frac{(-jqx)^{k}}{k!} dq $$

Since the complex exponential is entire, we can interchange the sum and integral, giving:

$$ I(t, 1) = \sum_{k=0}^{\infty} \frac{x^{k}}{k!} \int_{-\infty}^{\infty} \frac{(-jq)^{k}}{1-jq} dq $$

However, I get stuck here too because I can't find an antiderivative for the integrand.

Any ideas? Am I missing some fundamental theoretical concept or trick or technique that makes all of this difficulty disappear?

Unfortunately, I am really not that good at integration, but I want to get better!

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For simplicity we rename $j\to i, q \to -z $ to rewrite the integral as $$\begin{align} \int_{-\infty}^\infty\frac{e^{itz}dz}{(1+iz)^n}&=\int_\Gamma\frac{e^{itz}dz}{(1+iz)^n}=2\pi i\;\underset{z=i}{\text {Res}}\,\frac{e^{itz}}{(1+iz)^n}\\ &=\frac {2\pi}{ i^{n-1}}\left.\frac1 {(n-1)!}\frac {d^{n-1}}{dz^{n-1}} e^{itz}\right|_{z=i}=2\pi\frac{ t^{n-1}e^{-t}}{(n-1)!},\end{align} $$ where $\Gamma $ is the usual counterclockwise-oriented contour consisting of the real axis and a large semicircle in the upper complex half-plane. $t$ is assumed to be positive real number.

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  • $\begingroup$ This seems to solve it. Thank you very much. I need to revisit my contour integration! So what happens if t is a nonpositive real number? Also, if q = -z, then shouldnt dq = -dz ? But it doesnt matter because of the limits of integration, right? Sorry, for the basic questions, but the only way to get better is to ask when I'm unsure. $\endgroup$ – The Dude Feb 21 at 22:30
  • $\begingroup$ If $t $ is negative the coutour should lie in the lower half-plane, the pole is outside of the contour and the integral is 0. The negation of the integration variable and interchange of the limits always compensate each other. $\endgroup$ – user Feb 21 at 22:45
  • $\begingroup$ Thanks for your reply. I will have to go back to my notes and go over this. This was very helpful. $\endgroup$ – The Dude Feb 21 at 22:58
  • $\begingroup$ Okay, I spent a few days reviewing this and now have a question -- where does the $\frac{1}{i^{n-1}}$ come from? Shouldn't that be in the numerator when you take the derivative N-1 times? Why do you divide it out beforehand? $\endgroup$ – The Dude Feb 26 at 16:07
  • $\begingroup$ @TheDude $\frac1{(1+iz)^n}=\frac1{i^n (z-i)^n}$. $\endgroup$ – user Feb 26 at 16:29
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If we define the Fourier transform of a function $f$ and its inverse as $$\mathcal F(f)(\omega)=\int_{\mathbb R}f(x)e^{-j\omega x}dx \,\,\,\text{ and }\,\,\, f(x)=\frac 1 {2\pi}\int_{\mathbb R} \mathcal F(f)(\omega)e^{jx\omega}d\omega$$ then you're looking for the Fourier transform of $f_n(x)=f(x)^n$ where $f(x)=\frac{1}{1-jx}$.

Because the Fourier transform maps products to convolutions (times $\frac 1 {2\pi}$, given the Fourier definition we adopted), you're looking for the $n$-th self-convolution of $\mathcal F(f)$. Now, with $H$ denoting the Heaviside step function, we have $$f(x)=\frac{1}{1-jx}=\int_{\mathbb R}e^{-\omega}H(\omega)e^{jx\omega}d\omega$$ As a consequence, the Fourier transform of $f$ is $\omega\rightarrow 2\pi e^{-\omega}H(\omega)$.

This gives us $$I(t, 1)=2\pi e^{-t}H(t)$$ and you can verify that the $n$-th self-convolution is given by $$I(t,n)=2\pi\frac{t^{n-1}}{(n-1)!}e^{-t}H(t)$$

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  • $\begingroup$ Where did you find this fourier transform pair? After looking this up, I only found the first result. I'm not sure still how the (n-1)! comes out of this ,,Fourier Matching'' as I like to call it. $\endgroup$ – The Dude Feb 22 at 14:34
  • $\begingroup$ It's a known Fourier transform. Then computing the convolution is not too difficult. You can prove it by induction after trying the first few values of $n$. $\endgroup$ – Stefan Lafon Feb 22 at 14:36
  • $\begingroup$ Well I guess I am just a n00b. Okay, fair enough. I gotta practice more. $\endgroup$ – The Dude Feb 22 at 14:38
  • $\begingroup$ Though the factorial is still confusing... $\endgroup$ – The Dude Feb 22 at 14:38
  • $\begingroup$ It comes from integrating powers of $\omega$. $\endgroup$ – Stefan Lafon Feb 22 at 14:41

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