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.
Contour integration for $\int_0^\infty \frac{x^{\alpha -1}}{(x+\beta)(x+\gamma)}dx$
complex integrationcomplex-analysis
Related Solutions
We provide support for the main steps. Introducing $\mathrm{Log}(z)$, the branch with argument in $[0,2\pi)$ we integrate
$$f(z) = \exp((1/2)\mathrm{Log}(z)) \frac{\mathrm{Log}(z)}{(1+z)^2}$$
along a keyhole contour with the slit on the positive real axis. We get in the limit above the slit
$$K = \int_0^\infty \sqrt{x}\frac{\log(x)}{(1+x)^2} \; dx.$$
Below the slit we find
$$\int_\infty^0 \exp(\pi i) \sqrt{x} \frac{\log(x)+2\pi i}{(1+x)^2} \; dx = \int_0^\infty \sqrt{x} \frac{\log(x)+2\pi i}{(1+x)^2} \; dx \\ = K + 2\pi i \int_0^\infty \sqrt{x} \frac{1}{(1+x)^2} \; dx = K + 2\pi i J.$$
We have the two contributions separated into real and imaginary parts, so we just need to compute the residue at $z=-1$ of $f(z).$ Differentiating we find
$$\frac{1}{2z} \exp((1/2)\mathrm{Log}(z)) \mathrm{Log}(z) + \frac{1}{z} \exp((1/2)\mathrm{Log}(z)).$$
Evaluate at $z=-1$ to get
$$-\frac{1}{2} \exp((1/2)\pi i) (\pi i) - \exp((1/2)\pi i) = \frac{\pi}{2} - i.$$
With $K$ and $J$ real we collect the contributions to obtain
$$2K + 2\pi i J = 2\pi i \times \left(\frac{\pi}{2} - i\right) = \pi^2 i + 2\pi.$$
This means that
$$\bbox[5px,border:2px solid #00A000]{ K = \pi \quad\text{and}\quad J = \frac{\pi}{2}.}$$
The required ML estimates for the circular components go through with $\lim_{R\to\infty} 2\pi R \times \sqrt{R} \log(R)/(1+R)^2 = 0$ and $\lim_{\epsilon\to 0} 2\pi \epsilon \times \sqrt{\epsilon} \log(\epsilon) /(1+\epsilon)^2 = 0.$
For $\alpha, \beta \geq 0$ and $\gamma > 0$ let $$ f_{\alpha,\beta,\gamma} (z) = \frac{\exp \left(\mathrm{i} z \frac{z^2 - \alpha^2}{z^2 - \beta^2}\right)}{z^2 + \gamma^2} \, , \, z \in \mathbb{C} \setminus \{\pm\beta, \pm \mathrm{i} \gamma\} \, . $$ By symmetry we have $$ I (\alpha,\beta,\gamma) = \frac{1}{2} \int \limits_{-\infty}^\infty f_{\alpha,\beta,\gamma} (x) \, \mathrm{d} x \, .$$ Clearly, the integral exists for every combination of parameters and is bounded by $\frac{\pi}{2 \gamma}$ .
We can compute its value using the residue theorem. First note that we have $$ \operatorname{Res} (f_{\alpha,\beta,\gamma}, \mathrm{i} \gamma) = \frac{1}{2 \mathrm{i} \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) $$ and (by Jordan's lemma) $$ \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z = 0 \, ,$$ where $\Gamma_R$ is a semi-circle of radius $R > 0$ in the upper half-plane. A naive application of the residue theorem therefore yields $$ I (\alpha,\beta,\gamma) = \frac{1}{2} \left[ 2 \pi \mathrm{i} \operatorname{Res} (f_{\alpha,\beta,\gamma}, \mathrm{i} \gamma) - \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z \right]= \frac{\pi}{2 \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) \, .$$ While the result is indeed correct, this calculation is only valid for $\alpha = \beta$ . In this case we obtain (as already mentioned in the question) $I(\alpha,\alpha,\gamma) = \frac{\pi}{2 \gamma}\mathrm{e}^{- \gamma}$ .
If $\alpha \neq \beta$ , $f_{\alpha,\beta,\gamma}$ has essential singularities on the real axis, namely at $\pm \beta$ . Thus we need to deform our contour using small semi-circles $\gamma_\varepsilon (\pm \beta)$ of radius $\varepsilon > 0$ centred at these points and show that their contribution to the integral vanishes as $\varepsilon \to 0$ .
For simplicity, I will only discuss the case $\beta = 0$ in detail. We want to find $$ \int \limits_{\gamma_\varepsilon (0)} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z = \varepsilon \int \limits_0^\pi \frac{\mathrm{e}^{\mathrm{i} \left[\phi + \varepsilon \mathrm{e}^{\mathrm{i}\phi} - \alpha^2 \varepsilon^{-1} \mathrm{e}^{-\mathrm{i} \phi}\right]}}{\gamma^2+\varepsilon^2 \mathrm{e}^{2 \mathrm{i} \phi}} \, \mathrm{d} \phi = \frac{\varepsilon}{\gamma^2} \int \limits_0^\pi \mathrm{e}^{\mathrm{i} \left[\phi - \alpha^2 \varepsilon^{-1} \mathrm{e}^{-\mathrm{i} \phi}\right]} \left[1 + \mathcal{O} (\varepsilon) \right] \, \mathrm{d} \phi \, .$$ The leading-order term can actually be calculated analytically: for $z \in \mathbb{C}$ we have $$ \int \limits_0^\pi \mathrm{e}^{\mathrm{i} \left[\phi - z \mathrm{e}^{-\mathrm{i} \phi}\right]} \, \mathrm{d} \phi = \mathrm{i} \left[(2 \operatorname{Si}(z) - \pi) z + 2 \cos(z)\right] \, . $$ The sine integral $\operatorname{Si}$ satisfies $\lim_{x \to \infty} \operatorname{Si}(x) = \frac{\pi}{2}$, so we obtain $$ \int \limits_{\gamma_\varepsilon (0)} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z \sim \frac{\mathrm{i} \varepsilon}{\gamma^2} \left[(2 \operatorname{Si}(\alpha^2 \varepsilon^{-1}) - \pi) \alpha^2 \varepsilon^{-1} + 2 \cos(\alpha^2 \varepsilon^{-1})\right] \stackrel{\varepsilon \to 0}{\longrightarrow} 0 $$ as desired. Now we are allowed to apply the residue theorem , which yields \begin{align} I(\alpha,0,\gamma) &= \frac{1}{2} \left[ 2 \pi \mathrm{i} \operatorname{Res} (f_{\alpha,0,\gamma}, \mathrm{i} \gamma) - \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z - \lim_{\varepsilon \to 0} \int \limits_{\gamma_\varepsilon (0)} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z \right] \\ &= \frac{\pi}{2 \gamma} \exp \left(- \frac{\alpha^2 + \gamma^2}{\gamma}\right) \, . \end{align}
Almost the same calculation yields \begin{align} I(\alpha,\beta,\gamma) &= \frac{1}{2} \left[ 2 \pi \mathrm{i} \operatorname{Res} (f_{\alpha,\beta,\gamma}, \mathrm{i} \gamma) - \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z - \lim_{\varepsilon \to 0} \int \limits_{\gamma_\varepsilon (\beta)} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z - \lim_{\varepsilon \to 0} \int \limits_{\gamma_\varepsilon (-\beta)} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z \right] \\ &= \frac{\pi}{2 \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) \end{align} for $\beta > 0$ .
Combining these results we conclude that the integral is given by $$ I(\alpha,\beta,\gamma) = \frac{\pi}{2 \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) $$ for every $\alpha,\beta \geq 0$ and $\gamma > 0$ .
Best Answer
$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{res}}$(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: $$J=\pi\csc\pi\alpha\cdot(1-\alpha)\beta^{\alpha-2}$$