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I know that with the case

$$\int_0^{\infty}xe^{-ax}dx$$

You just substitute $u=ax$ and use the gamma function after so that it evaluates to $\frac{1}{a^2}$. However this clearly involves switching the bounds. So in this case, if $a$ were a positive imaginary number we would have our upper bound be $u=i\infty$. So I suppose my question is to show why

$$\int_0^{\infty}xe^{-ax}dx$$

Obeys the equality to $\frac{1}{a^2}$ for complex $a$.

$EDIT$

I know how to do Integration by parts and u substitution, that’s not the problem. The problem is what complex analysis backs up the fact that the integral obeys the identity

$$\int_0^{\infty}xe^{-ax}dx = \frac{1}{a^2}$$

For some complex $a$.

MichaelCatliMath
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    Technically the integral doesn't converge for purely imaginary $a$. The condition $\operatorname{Re}(a) > 0$ is required. – Ninad Munshi Jul 11 '20 at 00:22
  • Ok yea but you’d end up with limits involving $i\infty$, so that is my question. – MichaelCatliMath Jul 11 '20 at 00:22
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    You can see why substitution doesn't work well with complex numbers because you have to choose a path in the complex plane. The rigorous way of approaching this is creating a pie slice-shaped contour in the complex plane, with one leg of the pie on the real axis, the other leg at the angle of the complex number you want. The next step is showing that the integral on the arc is $0$, thus the integrals on the legs equal each other mod some constant. The heuristic of treating "$i\infty$" as a real $\infty$ works fine for the condition above, but you have to know what you're doing first. – Ninad Munshi Jul 11 '20 at 00:26
  • @Ninad Munshi Thank you so much. Is there any way you could write out the specifics? I just would like to see them if you don’t mind. I appreciate the comment. – MichaelCatliMath Jul 11 '20 at 05:27
  • @Harish Chandra Rajpoot I’m afraid not. In my question I pointed out why a substitution doesn’t particularly work. – MichaelCatliMath Jul 11 '20 at 05:29
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    Have a look at the details below. – Ninad Munshi Jul 11 '20 at 06:08

1 Answers1

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Consider the following paths in the complex plane

$$\begin{cases}\gamma_1(t) = t & t\in[0,R] \\ \gamma_2(t) = Re^{it} & t\in [0,-\arg(a)] \\ \gamma_3(t) = -te^{-i\arg(a)} & t\in [0,R]\end{cases}$$

for $\operatorname{Re}(a) > 0$ which form a closed, pie slice-shaped contour in the plane. The integrals, parametrized, equal

$$\int_0^R t e^{-at}\:dt + \int_0^{-\arg(a)}R^2ie^{-Rae^{it}+i2t}\:dt + \int_R^0 e^{-i2\arg(a)}te^{-|a|t}\:dt = 0$$

from complex analysis. The middle integral can be bounded

$$\left|\int_0^{-\arg(a)}R^2ie^{-Rae^{it}+i2t}\:dt\right| \leq |\arg(a)|R^2e^{-|\operatorname{Re}(a)|R} \to 0$$

as $R\to \infty$, thus we have that the integral we want equals a real valued integral times a complex constant

$$\int_0^\infty t e^{-at}\:dt = e^{-i2\arg(a)}\int_0^\infty te^{-|a|t}\:dt = \frac{e^{-i2\arg(a)}}{|a|^2} = \frac{1}{a^2}$$

Ninad Munshi
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