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I have seen $$\int_0^\infty e^{-ix}x^{s-1} \ \mathrm{d}x = i^{-s}\Gamma(s)$$ in a few posts regarding Mellin tranforms or a few difficult integrals. How can we prove this equality? I assume it can only be proven by residue theory.

  • It's better to use residue ideas. – openspace May 01 '20 at 19:35
  • I feel inclined to use the generalization $\int_0^{\infty} e^{-ax}x^{s-1}dx = a^{-s}\Gamma(s)$ but I did that with $u=ax$, and I’m sure the substitution $u=ix$ is invalid because $\lim_{x\to\infty} ix$ isn’t a valid thing.. don’t quote me on how I worded that though. Your desired answer comes right out of using that generalization, but I’m very unsure about the technicalities, or if it doesn’t include crucial steps validating $a=i$. – MichaelCatliMath May 01 '20 at 19:50
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    In fact, it's not true for at least real $s > 1$: the integral on the left diverges. You may need to be more careful about the allowed range of $s$. – The_Sympathizer May 01 '20 at 20:04
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    Does this answer your question? Can we find a closed form for this integral? – LL 3.14 May 01 '20 at 20:40

2 Answers2

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For $\Re(s) >0$ and $a > 0$ $$\int_0^\infty e^{-ax}x^{s-1} \ \mathrm{d}x = a^{-s}\Gamma(s)$$ By analytic continuation it stays true for $\Re(a) > 0,\Re(s) > 0$

and by continuity it holds for $\Re(a)\ge 0,a\ne 0,\Re(s)\in (0,1)$.

As usual the branch of $\log$ in $a^{-s}$ is implied by the derivation.

reuns
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  • I understand the idea of this generalized version but do you know of a mathematical proof regarding $a\in\mathbb{C}$? By the substitution ix=x in the latter integral we arrive at some integral in the form 0 to $i\infty$. How is the "direction" of $i $ impacting the infinity -- sorry this is a poorly worded question. –  May 01 '20 at 21:25
  • I don't see what you mean. Analytic continuation means that we do the substitution $y=ax$ in the integral only for $a > 0$, then we say that $\int_0^\infty e^{-ax}x^{s-1} \mathrm{d}x - a^{-s}\Gamma(s)$ is analytic for $\Re(a) > 0$ and it vanishes on $(0,\infty)$ so it vanishes for $\Re(a) > 0$ – reuns May 01 '20 at 21:43
  • If it is analytic for $\mathbb{R}(a)>0$, then how are we making the stipulation it is also analytic on $\mathbb{C}$? The original integral deals with $i$. –  May 01 '20 at 22:45
  • It is not analytic on $\Bbb{C}$, the integral converges for $\Re(a) \ge 0$. We deal with $\Re(a) > 0$ and at the last step we let $a\to i$ – reuns May 01 '20 at 22:49
  • Oh alright. I did not know we are allowed to do this. Do you have any references/ sources for your claim? Maybe I don't know all of the principles of analytic continuation aha. It seems interesting. Thank you. –  May 01 '20 at 23:26
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The integral does converge, if and only if $ s\in\left(0,1\right) $, in fact we need $ s $ to be less than $ 1 $ so that $ \lim\limits_{x\to +\infty}{x^{s-1}\,\mathrm{e}^{-\mathrm{i}x}}=0 $, and we need it to be greater than $ 0 $ so that the function can be integrated near $ 0 $.

Let $ s\in\left(0,1\right) $, and $ x\in\mathbb{R} \cdot $

Let's define a function $ f_{x} $ on $ \mathbb{R}_{+} $ by the following : $$ f_{x}:y\mapsto\frac{\mathrm{e}^{-\left(y^{\frac{1}{1-s}}+\mathrm{i}\right)x^{\frac{1}{1-s}}}}{y^{\frac{1}{1-s}}+\mathrm{i}} $$

Since $ f_{x} $ is continuous on $ \mathbb{R}_{+} $, and its absolute value can be upper-bounded by $ y\mapsto\frac{1}{y^{\frac{1}{1-s}}} $, it is integrable over $ \mathbb{R}_{+} $, thus, we can define a new function $ f $ on $ \mathbb{R} $ by : $$ f:x\mapsto\int_{0}^{+\infty}{f_{x}\left(y\right)\mathrm{d}y} $$

$ f $ is a $ \mathcal{C}^{1} $ function on $ \mathbb{R}_{+}^{*}=\left(0,+\infty\right) $, and using the dominated convergence theorem, we can prove that $ \lim\limits_{x\to +\infty}{f\left(x\right)}=0 $. We have that for any $ x\in\left(0,+\infty\right) $ : $$ f'\left(x\right)=-\frac{1}{1-s}x^{\frac{1}{1-s}-1}\mathrm{e}^{-\mathrm{i}x^{\frac{1}{1-s}}}\int_{0}^{+\infty}{\mathrm{e}^{-y^{\frac{1}{1-s}}x^{\frac{1}{1-s}}}\,\mathrm{d}y} $$

Which becomes, substituting $ \small\left\lbrace\begin{aligned}u&=y^{\frac{1}{1-s}}x^{\frac{1}{1-s}}\\ \mathrm{d}y&=\frac{1}{x}su^{-s}\,\mathrm{d}u\end{aligned}\right. $ : \begin{aligned} f'\left(x\right)&=-x^{\frac{1}{1-s}-2}\mathrm{e}^{-\mathrm{i}x^{\frac{1}{1-s}}}\int_{0}^{+\infty}{u^{-s}\mathrm{e}^{-u}\,\mathrm{d}y}\\&=-x^{\frac{1}{1-s}-2}\mathrm{e}^{-\mathrm{i}x^{\frac{1}{1-s}}}\Gamma\left(1-s\right) \end{aligned}

Integrating with respect to $ x $ from $ 0 $ to $ \infty $ gives : $$ \fbox{$\begin{array}{rcl}\displaystyle\int_{0}^{+\infty}{x^{\frac{1}{1-s}-2}\mathrm{e}^{-\mathrm{i}x^{\frac{1}{1-s}}}\,\mathrm{d}x}=\frac{1}{\Gamma\left(1-s\right)}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{x^{\frac{1}{1-s}}+\mathrm{i}}}\end{array}$} $$

Substituting $ \small\left\lbrace\begin{aligned}t&=x^{\frac{1}{s-1}}\\ \mathrm{d}t&=\frac{1}{1-s}x^{\frac{1}{1-s}-1}\,\mathrm{d}x\end{aligned}\right. $ in the LHS, then using Euler's reflection formula in the RHS, we get : $$ \fbox{$\begin{array}{rcl}\displaystyle\int_{0}^{+\infty}{t^{s-1}\mathrm{e}^{-\mathrm{i}t}\,\mathrm{d}t}=\frac{1}{\pi}\left(1-s\right)\Gamma\left(s\right)\sin{\left(\pi s\right)}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{x^{\frac{1}{1-s}}+\mathrm{i}}}\end{array}$} $$

I Guess I'll leave the rest to you, $ \int\limits_{0}^{+\infty}{\frac{\mathrm{d}x}{x^{\frac{1}{1-s}}+\mathrm{i}}}=\int\limits_{0}^{+\infty}{\frac{x^{\frac{1}{1-s}}-\mathrm{i}}{x^{\frac{2}{1-s}}+1}\,\mathrm{d}x} $, which is a particular case of some well known integral that can be computed using some complex analysis.

CHAMSI
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