How do I evaluate $\int_{0}^{\infty} u^{z-1}(e^{iu}-1) \, du$?

Question

I am trying to evaluate the following integral that shows up in this paper

http://arxiv.org/pdf/1103.4306v1.pdf

$I=\int_{0}^{\infty} u^{z-1}(e^{iu}-1)du= \Gamma(z)e^{\frac{iz\pi}{2}}$

for $-1<Re(z)<0$.

I have no idea what type of contour I can choose. Can someone please help me?

Thank you in advance.

Answer 1

Consider the quarter of circle $\Gamma$ of radius $M$ centred at the origin and lying in the first quadrant, thus made of the radius $0$-$M$ the arc $M$-$iM$ and the radius $iM$-$0$. Consider the function of the complex plane $\zeta=x+iy$ $$ f(\zeta) =\zeta^{z-1}\left( e^{i\zeta}-1\right); $$ choosing the logarithm branch cut in the negative $x$ axis, $f$ is holomorphic within $\Gamma$, thus $$ \oint_\Gamma f(\zeta) d\zeta=0 $$ hence $$ \int_0^M x^{z-1}\left(e^{ix}-1\right)dx + \int_0^{\pi/2} \left(M e^{i\varphi}\right)^{z-1} \left(e^{iMe^{i\varphi}}-z\right) iM e^{i\varphi}d\varphi -i^{z}\int_0^M y^{z-1}\left( e^{-y}-1\right)dy=0. $$ The contribution of the arc is under control, since letting $\Re (z)=\xi$, $\Im (z)=\eta$, $$ \left|\int_0^{\pi/2} \left(M e^{i\varphi}\right)^{z-1} \left(e^{iMe^{i\varphi}}-z\right) iM e^{i\varphi}d\phi\right| \le M^\xi \int_0^{\pi/2}e^{-\eta\varphi}\left| e^{iM\cos\varphi - M\sin\varphi}-1\right|d\varphi\\ \le 2M^\xi \int_0^{\pi/2}e^{-\eta\varphi}d\varphi=2M^{\xi}\frac{1-e^{-\eta\pi/2}}{\eta}\xrightarrow[M\to\infty]{}0, $$ since $\xi<0$. Terefore $$ \int_0^\infty x^{z-1}\left(e^{ix}-1\right)dx=i^{z}\int_0^\infty y^{z-1}\left( e^{-y}-1\right)dy=e^{i\pi z/2}\int_0^\infty y^{z-1}\left( e^{-y}-1\right)dy. $$ Integrating by parts $$ \int_0^\infty y^{z-1}\left( e^{-y}-1\right)dy= \frac{y^z}{z}\left(e^{-y}-1\right)\Big|_0^\infty+\frac{1}{z}\int_0^\infty y^z e^{-y}dy=\frac{1}{z}\Gamma(z+1)=\Gamma(z), $$ where we have used $$ \lim_{y\to\infty}\frac{y^z}{z}\left(e^{-y}-1\right)=0 $$ $$ \lim_{y\to0}\frac{y^z}{z}\left(e^{-y}-1\right)=0, $$ the second one following from the request $\xi>-1$. Putting everything back together, $$ \int_0^\infty x^{z-1}\left(e^{ix}-1\right)dx=e^{i\pi z/2}\Gamma(z) $$ if $-1<\Re (z)<0$.

Answer 2

You can use a quarter-circle in the first quadrant that is indented at the origin.

Integrating $f(s) = s^{z-1} (e^{is}-1)$ around the contour and using the fact that $-1 < \text{Re}(z) <0$, we get

$$\int_{0}^{\infty} u^{z-1} (e^{iu}-1) \, du + \int_{\infty}^{0} (e^{i \pi/2}t)^{z-1}(e^{-t}-1) e^{i \pi/2} \, dt =0. $$

So $$\int_{0}^{\infty} u^{z-1} (e^{iu}-1) \, du = e^{i \pi z /2}\int_{0}^{\infty} t^{z-1} (e^{-t}-1) \, dt \ .$$

Now consider the integral representation $$\Gamma(z) = \int_{0}^{\infty}t^{z-1} e^{-t} \, dt \ , \ \text{Re}(z) >0. \tag{1}$$

From $(1)$ subtract the Laplace transform $$\frac{\Gamma(z)}{a^{z}} = \int_{0}^{\infty} t^{z-1} e^{-at} \, dt \ , \ \text{Re}(z) >0, \text{Re}(a)>0$$ to get $$\Gamma(z) \left(1- a^{-z}\right) = \int_{0}^{\infty} t^{z-1} \left(e^{-t} - e^{-at} \right) \, dt \ , \ \text{Re}(z) > -1. $$

Then restrict $z$ to the vertical strip $-1 < \text{Re}(z) <0$ and let $a \to 0$.

So finally we get $$\int_{0}^{\infty} u^{z-1} (e^{iu}-1) \, du = \Gamma(z) e^{i \pi z/2} \ , \ -1<\text{Re}(z) <0 .$$