Many of us have seen the evaluation of the integral

$$\int^{\infty}_0 \frac{dx}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$

It can be solved using contour integration or beta function .

I thought of how to solve the integral

$$\int^{\infty}_0 \frac{\log(1+x)}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$

It can be solved using real methods as follows

consider the following integral

$$\int^{\infty}_0 x^{-p}(1+x)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)}$$

Differentiating with respect to $s$ we get

$$\int^{\infty}_0 x^{-p}(1+x)^{s-1}\log(1+x) dx=\frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)} \left(\psi_0 (1-s)- \psi_0(p-s)\right)$$

at $s =0$ we get

$$\int^{\infty}_0 x^{-p}\frac{\log(1+x)}{1+x} dx=\frac{\pi}{\sin(\pi p)} \left(\psi_0 (1)- \psi_0(p)\right)$$

where i used the reflection formula .

Statement of question

How to solve the following integral using contour integration

$$\int^{\infty}_0 \frac{\log(1+x)}{x^p(1+x)}\, dx \,\,\, 0<\Re(p)<1$$

I thought we can use the following contour

enter image description here

So the function

$$F(z) = \frac{e^{-p \log(z)}\log(1+z)}{(1+z)} $$

is analytic in and on the contour by choosing the branch cut of $e^{-p \log(z)}$ as $0\leq \text{Arg}(z)<2\pi$ and the branch cut of $\log(1+z)$ as $0\leq \text{Arg}(z+1)<2\pi$ so the function $F(z)$ is analytic everywhere except at $z\geq -1$ . I am finding difficulty finding the integral on the branch point $z=-1$ it seems there is a contribution of the branch point and the pole .

Please don't make any substitutions or simplifications for the integral. Feel free to use another contour if my choice was wrong .

  • $\begingroup$ I fail to see how residue theory will help here. About the branch point at $z=-1$, the integral diverges as $\log{\epsilon}$ as $\epsilon \to 0$, where $\epsilon$ is the radius of the arc about the branch point. Please correct me if I am mistaken. $\endgroup$ – Ron Gordon Jul 31 '13 at 17:10
  • $\begingroup$ yup Ron , this is what I thought . I think this will cancel with the integral on the interval [-1,0]. $\endgroup$ – Zaid Alyafeai Jul 31 '13 at 17:15
  • $\begingroup$ I see, so there is a factor of $\log{\epsilon}$ to carry around and hope cancels. BTW my idea was to approach the branch point from the other side of the contour (i.e., the negative real axis). $\endgroup$ – Ron Gordon Jul 31 '13 at 17:17
  • $\begingroup$ So you are thinking of using a semi-circle ? I thought of finding the integral with special case $p = \frac{1}{2}$ but that didn't work either . The problem started here integralsandseries.prophpbb.com/topic106.html. $\endgroup$ – Zaid Alyafeai Jul 31 '13 at 17:22
  • $\begingroup$ No, a full circle, but integrating along $(-\infty,-1)$. $\endgroup$ – Ron Gordon Jul 31 '13 at 17:23

To evaluate the integral using contour integration, consider the integral

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

where $C$ is the following contour:

enter image description here

The magnitude of the integral about the large arc of radius $R$ behaves as $\frac{\log{R}}{R^p}$ as $R \to \infty$ and thus vanishes. Let the radius of the small circular arcs be $\epsilon$. The contour integral is then equal to, in this limit,

$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} - e^{-i \pi p}\int_{\infty}^{1+\epsilon} dx \frac{x^{-p} [\log{(x-1)}+i \pi]}{1-x} \\ - e^{-i \pi p}\int_{1+\epsilon}^{\infty} dx \frac{x^{-p} [\log{(x-1)}-i \pi]}{1-x}+i \epsilon \int_{\pi}^{-\pi} d\phi \,e^{i \phi} \frac{(e^{i \pi}+\epsilon e^{i \phi})^{-p} \log{(\epsilon e^{i \phi})}}{\epsilon e^{i \phi}}$$

The first integral represents the integral about the branch cut along the positive real axis, which concerns the $z^{-p}$ term only. Note that the integral about the origin vanishes as $\epsilon \to 0$. The second and third integrals represent the integrals along each side of the branch cut on the negative axis. Note that, along this branch cut, the argument of $z^{-p}$ is $-\pi p$ on either side of the branch cut, as the branch cut there concerns the log term only. The fourth integral is the integral about the branch point $z=-1$.

By Cauchy's theorem, the contour integral is zero. Thus, we have

$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = i 2 \pi \, e^{-i \pi p} \int_{1+\epsilon}^{\infty} dx \frac{x^{-p}}{x-1} + i e^{-i \pi p} \int_{-\pi}^{\pi} d\phi \left ( \log{\epsilon} + i \phi \right )$$

Note that

$$\begin{align}\int_{1+\epsilon}^{\infty} dx \frac{x^{-p}}{x-1} &= \int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} + O(\epsilon) \end{align} $$


$$\begin{align}\int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} &= \int_0^{1-\epsilon} dx \, x^p \left (\frac1x + \frac1{1-x} \right ) \\ &= \frac1p (1-\epsilon)^p - \left [\log{(1-x)} x^p \right ]_0^{1-\epsilon} + p \int_0^{1-\epsilon} dx \, x^{p-1} \log{(1-x)}\\ &= \frac1p - \log{\epsilon} + p \int_0^1 dx \, x^{p-1} \log{(1-x)} + O(\epsilon)\\ &= \frac1p - \log{\epsilon} - p \sum_{k=1}^{\infty} \frac1{k (k+p)}+ O(\epsilon) \\ &= \frac1p - \log{\epsilon}-\gamma -\psi(1+p)+ O(\epsilon) \\ &= - \log{\epsilon} - (\gamma + \psi(p))+ O(\epsilon) \end{align} $$

where $\psi$ is the digamma function. Note that the singular $\log{\epsilon}$ pieces cancel. The second piece of the second integral on the RHS vanishes as it is an odd function over a symmetric interval. Thus, we may take the limit as $\epsilon \to 0$ and we get

$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = -\frac{i 2 \pi \, e^{-i \pi p}}{1-e^{-i 2 \pi p}} (\gamma + \psi(p)) = -\frac{\pi}{\sin{\pi p}} (\gamma + \psi(p))$$

Alternatively, we may express this so that it is clear that the integral takes a positive value:

$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = \frac{\pi}{\sin{\pi p}} \left (\frac1p - H_p \right ) $$

where $H_p$ is the analytically continued harmonic number at $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.