# How to evaluate the integral $\int_{0}^{\infty} \frac{\log x}{x^p (x-1)}\, dx$?

How to evaluate the integral $$I_p=\int_{0}^{\infty} \frac{\log x}{x^p (x-1)}\, dx\,?$$

Attempt:

My first idea was to find the area of convergence, which gives me $$p \in (0,1)$$. Now let's consider the function $$f(z) = \frac{\log^2 z}{z^p (z-1)}$$. Then we can use Cauchy theorem and residuals.

But I'm stuck here. Because there is no term to describe $$I_1$$. Any ideas?

We can take from here the following representation of the digamma function: $$\psi(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}dx\Rightarrow \psi_1(s)=\int_0^1 \frac{x^{s-1}\ln x}{x-1}dx$$ Now we are just one step away to solve it by splitting the integral into two pieces because we have: $$\int_1^\infty \frac{\ln x}{x^p(x-1)}dx\overset{\large x=\frac{1}{t}}=\int_0^1 \frac{t^{p-1}\ln t}{t-1}dt$$

$$\int_0^\infty \frac{\ln x}{x^p(x-1)}dx=\int_0^1 \frac{x^{-p}\ln x}{x-1}dx+\int_0^1\frac{x^{p-1}\ln x}{x-1}dx=\psi_1(1-p)+\psi_1(p)=\boxed{\frac{\pi^2}{\sin^2 (\pi p)}}$$ Above follows by the the trigamma's reflection formula.

• Nice solution. I posted one in which I avoid any reference to special functions and instead uses contour integration and makes use of the partial fraction expansion of the cotangent function. Let me know what you think. – Mark Viola Mar 18 at 3:26
• @MarkViola that is cute and probably what OP intended to find (seeing the complex analysis tag)! I wish I knew to do that myself from scratch (I don't have a good knowledge of complex analysis). – Zacky Mar 18 at 9:47
• Thank you. And I thought it might be of interest to you to see THESE derivations of the partial fraction representation of the cotangent function, which was key to my solution herein. – Mark Viola Mar 18 at 16:45

Well use $$\Re s>0\implies \int_0^\infty\frac{ue^{-su}}{1-e^{-u}}du=\sum_{n\ge 0}\frac{1}{(n+s)^2}.$$With the substitution $$u=\ln x$$ followed by the identity $$\int_{\Bbb R}g(u)du=\int_0^\infty [g(u)+g(-u)]du$$, $$I_1=\int_0^\infty\frac{u (e^{-pu}+e^{-(1-p)u})}{1-e^{-u}}du=\sum_{n\ge 0}\left(\frac{1}{(n+1-p)^2}+\frac{1}{(n+p)^2}\right).$$ This is evaluated here as $$\pi^2\csc^2\pi p.$$

• That's beautiful! Thanks for your answer, I was quite interested in this problem.+1 – Awe Kumar Jha Mar 17 at 17:02
• Using the relation: $\zeta(2, x)=\psi_1(x)$ further simplification is avaible using the reflection formula for the trigamma function. I didn't find that formula for the Hurwitz Zeta function (on google), altough there must exist one. – Zacky Mar 17 at 18:22
• @Zacky Thanks; fixed. – J.G. Mar 17 at 18:26
• @J.G. Nice solution. I posted one in which I avoid any reference to special functions and instead uses contour integration and makes use of the partial fraction expansion of the cotangent function. Let me know what you think. – Mark Viola Mar 18 at 3:27
• Your solution is fine; you might consider providing readers with a reference to your first equation since this is the thrust of the development. And I just thought it might be of interest to you to see THESE derivations of the partial fraction representation of the cotangent function. – Mark Viola Mar 18 at 16:42

The OP stated at the end of the posted question

Now let's consider the function $$f(z) = \frac{\log^2 z}{z^p (z-1)}$$. Then we can use Cauchy theorem and residuals.

Rather than analyze a contour integral of $$f(z)=\frac{\log^2 z}{z^p (z-1)}$$, we analyze the integral

$$I(p)=\oint_C \frac{\log(z)}{z^p(z-1)}\,dz$$

where $$C$$ is the classical keyhole contour where the branch cut for $$\log(z)$$ is take along the positive real axis and $$\arg(z)\in [0,2\pi)$$. Inasmuch as $$\frac{\log(z)}{z^p(z-1)}$$ is analytic in and on $$C$$, Cauchy's Integral Theorem guarantees that

\begin{align} 0&=\oint_C \frac{\log(z)}{z^p(z-1)}\,dz\\\\ &=\int_\epsilon^R \frac{\log(x)}{x^p(x-1)}\,dx+\underbrace{\int_0^{2\pi }\frac{\log(Re^{i\phi})}{R^pe^{ip\phi}(Re^{i\phi}-1)}\,iRe^{i\phi}\,d\phi}_{\to 0\,\,\text{as}\,\,R\to \infty}+\int_R^{1+\nu}\frac{\log(x)+i2\pi}{x^pe^{i2\pi p}(x-1)}\,dx\\\\ &+\underbrace{\int_{2\pi}^\pi \frac{\log(1+\nu e^{i\phi})}{(1+\nu e^{i\phi})^p\nu e^{i\phi}}\,i\nu e^{i\phi}\,d\phi}_{\to -i\pi (i2\pi )e^{-i2\pi p}\,\,\text{as}\,\,\nu\to0}+\int_{1-\nu}^\epsilon \frac{\log(x)+i2\pi}{x^pe^{i2\pi p}(x-1)}\,dx+\underbrace{\int_{2\pi}^0 \frac{\log(\epsilon e^{i\phi})}{(\epsilon e^{i\phi})^p(\epsilon e^{i\phi}-1)}\,i\epsilon e^{i\phi}\,d\phi}_{\to 0\,\,\text{as}\,\,\epsilon\to 0}\tag1 \end{align}

Letting $$\epsilon\to 0^+$$, $$R\to \infty$$, and $$\nu \to 0^+$$ in $$(1)$$, we obtain

$$(1-e^{-i2\pi p})\int_0^\infty \frac{\log(x)}{x^p(x-1)}\,dx=i2\pi e^{-i2\pi p}\,\,\text{PV}\left(\int_0^\infty \frac{1}{x^p(x-1)}\,dx\right)+(i\pi)\left(i2\pi e^{-i2\pi p} \right)\tag2$$

where $$\text{PV}\left(\int_0^\infty \frac{1}{x^p(x-1)}\,dx+i\pi\right)$$ denotes the Cauchy Principal Value (CPV) integral.

We can evaluate the CVP integral as follows.

\begin{align} \text{PV}\int_0^\infty \frac{1}{x^p(x-1)}\,dx&=\lim_{\nu\to 0^+}\left(\int_0^{1-\nu}\frac{1}{x^p(x-1)}\,dx+\int_{1+\nu}^\infty\frac{1}{x^p(x-1)}\,dx\right)\\\\ &=\lim_{\nu\to 0^+}\left(-\sum_{n=0}^\infty\int_0^{1-\nu}x^{n-p}\,dx+\sum_{n=0}^\infty\int_{1+\nu}^\infty x^{-n-p-1}\,dx\right)\\\\ &=\sum_{n=0}^\infty \left(\frac1{n+p}-\frac1{n-p+1}\right)\tag3 \end{align}

The series on the right-hand side of $$(3)$$ is the partial fraction representation of $$\pi \cot(\pi p)$$ for $$p\in (0,1)$$ (SEE HERE).

Substituting $$(3)$$ into $$(2)$$, dividing both sides by $$(1-e^{-i2\pi p})$$, we find

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

• @openspace Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. – Mark Viola Mar 25 at 17:55

# Too Long for comment

If $$p≥2$$ were some natural number, we could write the indefinite integral as, $$I = \log x \left[ \log \left( 1-\frac {1}{x} \right) + \sum_{r=2}^p \frac {1}{(r-1)x^{r-1}} \right] - \text {Li}{ }_2 \left( \frac {1}{x} \right) + \sum_{r=2}^p \frac {1}{(r-1)^2x^{r-1}} + C$$ However, I don't know how this can be related to your problem, as you say $$0 which seems absurd.