I've stumbled upon one integral which is rather challenging because of the fractional power of t:

\begin{equation}\int_0^\infty \log(1+tx)t^{-p-1}dt, \end{equation} where $p\in(0,1)$ and $x>0$. Any idea how to approach it using contour integration?

Update: to add some interesting background, it turns out that this integral is used to express a very simple expression \begin{equation} x^p = \frac{1}{\pi}p \sin(p\pi)\int_0^\infty \log(1+tx)t^{-p-1}dt \end{equation}


I know this isn't contour integration, but a proof of your expression using Feynman integration - let

$$ I(x) = \int_0^{\infty} \log(1+tx)t^{-p-1} \ \mathrm{d}t, $$ $$ I'(x) = \int_0^{\infty} \frac{t^{-p}}{1+tx} \ \mathrm{d}t, $$ making the substitution $u=tx$, $$ I'(x) = x^{p-1}\int_0^{\infty} \frac{u^{-p}}{1+u} \ \mathrm{d}u. $$ Recall the result that $$ \int_0^{\infty} G(u)f(u) \ \mathrm{d}u = \int_0^{\infty} F(u)g(u) \ \mathrm{d}u, $$ where $F$ is the Laplace transform of $f$ and $G$ is the Laplace transform of $g$. Let $$F(s) = \frac{1}{1+s}, \quad f(t) = \mathrm{e}^{-t}$$ and $$g(t) = t^{-p}, \quad G(s) = \frac{\Gamma(1-p)}{s^{1-p}},$$ then $$ I'(x) = x^{p-1}\Gamma(1-p)\int_0^{\infty} u^{p-1}\mathrm{e}^{-u} \ \mathrm{d}u. $$ The remaining integral is the definition of $\Gamma(p)$. Recall the reflection formula, $$\Gamma(1-p)\Gamma(p) = \frac{\pi}{\sin{(\pi p)}},$$


$$ I'(x) = \frac{\pi x^{p-1}}{\sin{(\pi p)}}, $$

$$ I(x) = \frac{\pi x^{p}}{p \sin{(\pi p)}} + C, $$

where we can use the fact that $I(0) = 0$ to say that $C=0$. Hence $$ \int_0^{\infty} \log(1+tx)t^{-p-1} \ \mathrm{d}t = \frac{\pi x^{p}}{p \sin{(\pi p)}} $$ as claimed.

  • 1
    $\begingroup$ Brilliant approach. $\endgroup$ – Alex Lomachenko Oct 9 '15 at 10:21

This is a great problem for contour integration. Just tricky enough to be really interesting. What makes it interesting is that there are two functions in the integrand needing their own separate branch cuts. One must keep in mind that each function only needs one cut.

So to compute the integral, consider the following contour integral:

$$\oint_C dz \, \log{(1+x z)}\, z^{-p-1} $$

where $C$ is the following contour:

enter image description here

Note the two cuts. The cut along the positive real axis is for the $z^{-p}$ term, while the cut along the negative real axis is for the log term. We define the argument of $z$ to be $0$ above the positive cut and $2 \pi$ below, and $+ \pi$ above the negative cut and $- \pi$ below. I will omit the steps involving showing that the integrals about the small and large arcs vanish as their respective radii vanish and go to $\infty$.

Thus, the contour integral is

$$\left (1-e^{-i 2 \pi p} \right )\int_0^{\infty} dt \, \log{(1+x t)} \, t^{-p-1} \\+e^{-i \pi p} \int_{\infty}^{1/x} dt \, \left [\log{(x t-1)}+i \pi \right ] \, t^{-p-1}+ e^{-i \pi p}\int_{1/x}^{\infty} dt \, \left [\log{(x t-1)}-i \pi \right ] \, t^{-p-1}$$

Note that I have not invoked the branch cut for the power piece of the integrand. Rather, I have set the argument of $z$ to be $\pi$ there as I would without the log piece. This is the tricky part of doing the complex analysis.

By Cauchy's theorem, the contour integral is zero in the above-mentioned limit. Thus,

$$\left (1-e^{-i 2 \pi p} \right )\int_0^{\infty} dt \, \log{(1+x t)} \, t^{-p-1} = i 2 \pi e^{-i \pi p} \int_{1/x}^{\infty} dt \, t^{-p-1} = i 2 \pi e^{-i \pi p} \frac{x^p}{p} $$

The stated result follows:

$$\int_0^{\infty} dt \, \log{(1+x t)} \, t^{-p-1} = \frac{\pi \, x^p}{p \sin{\pi p}} $$

  • $\begingroup$ nice work as always, Ron. $\endgroup$ – Bennett Gardiner Oct 15 '15 at 15:20

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