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I've shown that for all $p \in \mathbb{R}^{*+}$ $$\int_{0}^{+\infty}e^{-x^p}\text{d}x=\Gamma\left(1+\frac{1}{p}\right)$$ And I want to show that $$ \int_{0}^{+\infty}e^{ix^n}\text{d}x=\Gamma\left(1+\frac{1}{n}\right)e^{i\pi/2n} $$ Is that possible with a simple change of variable ?

Martin Sleziak
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Atmos
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  • Yes, you can use a change of variable (specifically, a rotation of $x$ in the complex plane). – Michael L. Jan 15 '18 at 23:20
  • Note that complex changes of variable are intimately tied to holomorphicity. You aren't guaranteed to get the same result without it. – Cameron L. Williams Jan 15 '18 at 23:25
  • You need to use some care, since $$\int_{0}^{+\infty}e^{ix^n}\text{d}x$$ does not exist as a Lebesgue integral, while $$\int_{0}^{+\infty}e^{-x^p}\text{d}x$$ does. – GEdgar Jan 16 '18 at 00:48

1 Answers1

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$x=u^{1/n}$, $dx=\frac1n u^{1/n-1}\,du$

$$\begin{align} \int_0^{\infty} e^{ix^n}\,dx&\overbrace{=}^{x\mapsto u^{1/n}}\frac1n \int_0^\infty e^{iu}u^{1/n-1}\,du\\\\ &\overbrace{=}^{u\mapsto ix}\frac1n \int_0^{i\infty} e^{-x} (ix)^{1/n-1}\,i\,dx\\\\ &=\frac1n e^{i\pi/(2n)}\int_0^{i\infty} x^{1/n-1}e^{-x}\,dx\tag1\\\\ &=\frac1n e^{i\pi/(2n)}\int_0^{\infty} x^{1/n-1}e^{-x}\,dx\tag2\\\\ &=e^{i\pi/(2n)}\frac1n \Gamma\left(\frac1n\right)\tag3\\\\ &=e^{i\pi/(2n)}\Gamma\left(1+\frac1n\right)\tag 4 \end{align}$$

where in going from $(1)$ to $(2)$, we used Cauchy's Integral Theorem to deform the contour back to the real line, and in going from $(3)$ to $(4)$ we used the function equation $\Gamma(1+x)=x\Gamma(x)$.

Note in using Cauchy's Integral Theorem to go from $(1)$ to $(2)$, we need to account for a branch point at $z=0$ in the complex plane. We chose to cut the plane along the negative real axis.

Then, we can write

$$\begin{align} 0&=\oint_C e^{-z} z^{1-1/n}\,dz\\\\ &=\int_{i\epsilon}^{iR} x^{1/n-1}e^{-x}\,dx-\int_{\epsilon}^R x^{1/n-1}e^{-x}\,dx\\\\ &+\int_0^{\pi/2}(\epsilon e^{i\phi})^{1/n-1}e^{-\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi-\int_0^{\pi/2}(R e^{i\phi})^{1/n-1}e^{-R e^{i\phi}}\,iR e^{i\phi}\,d\phi\tag5 \end{align}$$

where we exploited the fact that $z^{1/n-1}e^{-z}$ is analytic in and on $C$ (i.e., We have excluded the branch point and corresponding cut from $C$).

It is straightforward to show that the third and fourth integrals on the right-hand side of $(5)$ vanish as $\epsilon\to 0$ and $R\to \infty$.

Hence , we deduce the coveted relationship

$$\int_{0}^{I\infty} x^{1/n-1}e^{-x}\,dx=\int_{0}^\infty x^{1/n-1}e^{-x}\,dx$$

as was to be shown.

Mark Viola
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