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

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!

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.

• 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. – The Dude Feb 21 at 22:30
• 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. – user Feb 21 at 22:45
• Thanks for your reply. I will have to go back to my notes and go over this. This was very helpful. – The Dude Feb 21 at 22:58
• 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? – The Dude Feb 26 at 16:07
• @TheDude $\frac1{(1+iz)^n}=\frac1{i^n (z-i)^n}$. – user Feb 26 at 16:29

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 $$I(t,n)=2\pi\frac{t^{n-1}}{(n-1)!}e^{-t}H(t)$$

• 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. – The Dude Feb 22 at 14:34
• 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 $n$. – Stefan Lafon Feb 22 at 14:36
• Well I guess I am just a n00b. Okay, fair enough. I gotta practice more. – The Dude Feb 22 at 14:38
• Though the factorial is still confusing... – The Dude Feb 22 at 14:38
• It comes from integrating powers of $\omega$. – Stefan Lafon Feb 22 at 14:41