I know that it should be solved by integrating on a contour of two semicircles with radius $\epsilon$ and $T$, and the real line. Then as $\epsilon\to0$ and $T\to\infty$ it should be the value of residue. How can I calculate the residue?

Note: I checked with wolframalpha, and it should be $$-\cos\left(\frac{\pi\alpha}{2}\right)\Gamma\left(-\alpha\right).$$

  • $\begingroup$ You need to decide of what function you should compute the residue. Did you? $\endgroup$ – John B May 1 '16 at 16:20
  • $\begingroup$ I think that it should be $\frac{1-e^{-x}}{x^{\alpha+1}}$. $\endgroup$ – Hodossy Szabolcs May 1 '16 at 16:42

Let $I(\alpha)$ be the integral

$$I(\alpha)=2\alpha\int_0^\infty \frac{1-\cos(x)}{x^{1+\alpha}}\,dx \tag 1$$

for $0<\alpha<2$. Integrating $(1)$ by parts with $u=1-\cos(x)$ and $v=-\frac{1}{\alpha x^{\alpha}}$ reveals

$$I(\alpha)=2\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx \tag 2$$

Now, moving to the complex plane, we analyze the closed contour integral

$$J(\alpha)=\oint_C \frac{e^{iz}}{z^\alpha}\,dz \tag 3$$

where $C$ is comprised of (i) the segment on the real line from $\epsilon$ to $R$, (ii) the quarter circle with radius $R$, centered at the origin, from $R$ to $iR$, (iii) the line segment along the imaginary axis from $iR$ to $i\epsilon$, and (iv) the quarter circle with radius $\epsilon$, centered at the origin, from $i\epsilon$ to $\epsilon$.

With the branch cut chosen as the line along the non-positive real axis, from $0$ to $-\infty$, the integrand in $3$ is analytic in and on $C$. Then, Cauchy's Integral Theorem guarantees that $J(\alpha)=0$.

In the limit as $R\to \infty$ and $\epsilon \to 0$, the contributions to $J(\alpha)$ from the integrations along the quarter circles vanish. Hence, we find that

$$\begin{align} 2\int_0^\infty \frac{e^{ix}}{x^\alpha}\,dx&=2 (i)^{1-\alpha}\int_0^\infty \frac{e^{-y}}{y^{\alpha}}\,dy\\\\ &=2\,e^{i\pi(1-\alpha)/2}\int_0^\infty y^{-\alpha}e^{-y}\,dy\\\\ &=2\,e^{i\pi(1-\alpha)/2}\Gamma(1-\alpha)\tag 4 \end{align}$$

Taking the imaginary part of both sides of $(4)$ reveals that $(2)$ becomes

$$I(\alpha)=2\cos(\alpha \pi/2)\Gamma(1-\alpha)$$

Using the functional equation $\Gamma (1+x)=x\Gamma(x)$ yields

$$\bbox[5px,border:2px solid #C0A000]{2\alpha \int_0^\infty \frac{1-\cos(x)}{x^{1+\alpha}}\,dx=-2\alpha \cos(\alpha \pi/2)\Gamma(-\alpha)}$$

as was to be shown!


We have: $$ I(\alpha) = 2\alpha\int_{0}^{+\infty}\frac{1-\cos x}{x^{\alpha+1}}\,dx \stackrel{IBP}{=} 2\int_{0}^{+\infty}\frac{\sin x}{x^{\alpha}}\,dx$$ but since $\mathcal{L}(\sin x) = \frac{1}{1+s^2}$ and $\mathcal{L}^{-1}(x^{-\alpha})=\frac{s^{\alpha-1}}{\Gamma(\alpha)}$ we also have: $$ I(\alpha) = \frac{2}{\Gamma(\alpha)}\int_{0}^{+\infty}\frac{s^{\alpha-1}}{s^2+1}\,ds $$ and the last integral is a value of the Euler beta function.

By the reflection formula for the $\Gamma$ function the claim follows.


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