# Contour integration for $\int_0^\infty \frac{x^{\alpha -1}}{(x+\beta)(x+\gamma)}dx$

complex integrationcomplex-analysis

I have been trying to compute the following integral using contour integration and the residue theorem for quite some time now and can't get it to work out:
$$\int_0^\infty \frac{x^{\alpha -1}}{(x+\beta)(x+\gamma)}dx$$
for $$0<\alpha<1$$ and $$\gamma,\beta>0$$. I assume that we consider the function $$f(z) = \frac{z^{\alpha -1}}{(z+\beta)(z+\gamma)}$$, where $$z^{\alpha -1}= e^{(\alpha-1)\log z}$$. What should we choose our contour to be? I have tried things like a keyhole contour but they don't seem to work.

$$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{res}}$$(to remove this from "unanswered")

Let $$\log:\Bbb C^\star\to\Bbb C$$ be the logarithm defined by $$0\le\arg<2\pi$$. Let $$0<\alpha<\color{red}{2}$$ and $$\beta,\gamma>0$$. On $$\Bbb C\setminus[0,\infty)$$, we have a meromorphic function $$f:z\mapsto(z+\beta)^{-1}(z+\gamma)^{-1}\exp((\alpha-1)\log z)$$. From now on, read $$z^{\alpha-1}$$ as $$\exp((\alpha-1)\log z)$$.

The case $$\alpha=1$$ is to be momentarily discarded, then once we have a result we take limits as $$\alpha\to1$$. The final results must be understood as limits for $$\alpha=1$$.

The standard keyhole contour works. Observe that $$|zf(z)|$$ is asymptotically equivalent to $$|z|^{\alpha-2}$$ as $$|z|\to\infty$$, which vanishes uniformly. Therefore the large arc vanishes (by the estimation lemma and some basic analysis, no explicit bounds required). Similarly, $$|zf(z)|$$ is asymptotically equivalent to $$\beta^{-1}\gamma^{-1}\cdot|z|^\alpha$$ as $$|z|\to0$$, which vanishes uniformly. Therefore the small arc vanishes.

Fix $$x>0$$. As $$z\to x$$ in the upper half plane: $$f(z)\to\frac{x^{\alpha-1}}{(x+\beta)(x+\gamma)}$$As $$z\to x$$ in the lower half plane: $$f(z)\to e^{2\pi i\alpha}\cdot\frac{x^{\alpha-1}}{(x+\beta)(x+\gamma)}$$

Assume $$\beta\neq\gamma$$.

We conclude that: $$(1-e^{2\pi i\alpha})\cdot J=2\pi i\cdot(\res(f;-\beta)+\res(f;-\gamma))$$ Where $$J$$ is your integral.

Then all poles are simple and we can find the residues by inspection: $$J=-\pi\cdot e^{-\pi i\alpha}\cdot\frac{2i}{e^{\pi i\alpha}-e^{-\pi i\alpha}}\cdot\left(\frac{\beta^{\alpha-1}\cdot e^{\pi i\alpha-\pi i}}{\gamma-\beta}+\frac{\gamma^{\alpha-1}\cdot e^{\pi i\alpha-\pi i}}{\beta-\gamma}\right)$$Which simplifies to: $$J=\pi\csc\pi\alpha\cdot\frac{\gamma^{\alpha-1}-\beta^{\alpha-1}}{\beta-\gamma}$$

Now assume $$\beta=\gamma$$. Then: $$(1-e^{2\pi i\alpha})\cdot J=2\pi i\cdot\res(f;-\beta)$$The residue will be the first order term in the Taylor expansion of $$z\mapsto z^{\alpha-1}$$ at $$-\beta$$, which is $$e^{\pi i\alpha}(\alpha-1)\cdot\beta^{\alpha-2}$$. We find: $$J=\pi\csc\pi\alpha\cdot(1-\alpha)\beta^{\alpha-2}$$