Evaluate $\int_0^{\infty}\frac{\ln x}{x^a(x+1)}dx$ where $0<a<1$

Question

I'm trying to compute this integral, $$\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx \hbox{ where } 0<a<1$$ I drew a typical Pacman contour with branch cut at positive real axis. Then, we have $$\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz=\left(\int_{L_{1}}+\int_{L_{2}}+\int_{C_{R}}+\int_{C_{\epsilon}}\right)\frac{\ln z}{z^{a}(z+1)}dz$$ The function has a pole at $z=-1$. Then, by residue theorem, the whole integral becomes \begin{align*} \int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz&=2\pi i\text{res}(\frac{\ln z}{z^{a}(z+1)};-1)\\&=2\pi i\lim_{z\rightarrow-1}(z+1)\frac{\ln z}{z^{a}(z+1)}\\&=2\pi i\lim_{z\rightarrow-1}\frac{\ln z}{z^{a}}\\&=2\pi i\frac{\ln e^{i\pi}}{e^{i\pi a}}\\&=2\pi i\frac{\ln1+i\pi}{e^{i\pi a}}\\&=-\frac{2\pi^{2}}{(e^{i\pi})^{a}} \end{align*} On $L_1$, $z=xe^{i\epsilon}$ where $\epsilon\leq x\leq R$

On $L_{2}$, $z=xe^{i(2\pi-\epsilon)}$ where $\epsilon\leq x\leq R$

On $C_{R}$, $z=Re^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$

Lastly on On $C_{\epsilon}$, $z=\epsilon e^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$

So I tried to compute them separately and I anticipate the curves will both go to $0$ by ML inequality, and two linear integrals will consist of the term $\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$ and hence simply solve for it. But I'm having trouble physically solving it out. I think the log is throwing me off. Can anybody please help me?

Answer 1

The integral can be evaluated to $$\pi^2\cot\pi a~\csc\pi a$$

I will provide a proof later.

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Let $$J(a)=\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$$ $$I(a)=\int_0^\infty\frac{x^{-a}}{x+1}dx$$

Clearly, $J(a)=-I'(a)$.

Evaluation of $I(a)$ is easier.

Let $$f(z)=\frac{z^{-a}}{z+1}=\frac{\exp(-a(\ln|z|+i\arg z))}{z+1}\hbox{ where }\arg z\in[0,2\pi).$$

Let $C$ be the keyhole contour centered at the origin, avoiding the branch cut of $z^{-a}$.

By residue theorem, $$\oint_C f(z)dz=2\pi i\operatorname*{Res}_{z=-1}f(z)=2\pi i(-1)^{-a}=2\pi ie^{-\pi i a}\qquad{(1)}$$

Also, $$\oint_C =\int_{\text{large circle}}+\int_{\text{small circle}}+\int_{\text{upper real axis}}+\int_{\text{lower real axis}}$$

You can easily prove that the first two integrals tend to zero.

Moreover, $$\int_{\text{upper real axis}}=\int^\infty_0 f(te^{i0})dt=I(a)$$ $$\begin{align} \int_{\text{lower real axis}} &=\int_\infty^0 f(te^{i2\pi})dt \\ &=\int_\infty^0\frac{(te^{2\pi i})^{-a}}{t+1}dt \\ &=-e^{-2\pi i a}\int^\infty_0\frac{t^{-a}}{t+1}dt \\ &=-e^{-2\pi i a}I(a) \\ \end{align} $$

Back to $(1)$, $$I(a)-e^{-2\pi i a}I(a)=2\pi ie^{-\pi i a}$$ $$(e^{\pi i a}-e^{-\pi i a})I(a)=2\pi i$$ $$\frac{e^{\pi i a}-e^{-\pi i a}}{2i}I(a)=\pi$$ $$(\sin{\pi a})I(a)=\pi$$ $$I(a)=\pi\csc{\pi a}$$

Therefore, $J(a)=-I'(a)=\pi^2\cot\pi a~\csc\pi a$. $$\color{red}{\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx=\pi^2\cot\pi a~\csc\pi a}$$

Some special values are $$J(1/6)=2\sqrt3\pi^2, ~J(1/4)=\sqrt2\pi^2, ~J(1/3)=2\pi^2/3, ~J(1/2)=0$$

$J(a)$ satisfies the functional equation $$J(a)=-J(1-a).$$

Answer 2

Actually you do not strictly need Complex Analysis.
Using the Beta function and the reflection formula for the $\Gamma$ function one gets $$ \int_{0}^{+\infty}\frac{x^\beta}{x+1}\,dx = -\frac{\pi}{\sin(\pi\beta)}$$ for any $\beta\in(-1,0)$. By differentiating both sides with respect to $\beta$ it follows that $$ \int_{0}^{+\infty}\frac{x^\beta\log x}{x+1}\,dx = \frac{\pi^2\cos(\pi\beta)}{\sin^2(\pi\beta)}$$ so $$ \forall \alpha\in(0,1),\qquad \int_{0}^{+\infty}\frac{\log x}{x^\alpha(x+1)}\,dx = \color{red}{\frac{\pi^2\cos(\pi\alpha)}{\sin^2(\pi\alpha)}}.$$

Answer 3

We seek to compute using contour integration the integral

$$J = \int_0^\infty \frac{\ln x}{x^a (1+x)} \; dx$$

where $0\lt a\lt 1.$ We work with

$$f(z) = \frac{\mathrm{Log}(z)} {\exp(a\mathrm{Log}(z)) (1+z)}$$

where $\mathrm{Log}(z)$ is the branch with argument in $[0,2\pi).$ We use a keyhole contour wih radius $R$ and the slot on the positive real axis. Let $\Gamma_0$ be the segment on the real axis up to $R$, $\Gamma_1$ the big circle of radius $R$, $\Gamma_2$ the segment below the real axis coming in from $R$ and finally $\Gamma_3$ the small circle of radius $\epsilon$ enclosing the origin. We then have

$$\left(\int_{\Gamma_0} + \int_{\Gamma_1}+ \int_{\Gamma_2}+ \int_{\Gamma_3}\right) f(z) \; dz = 2\pi i \times \mathrm{Res}_{z=-1} f(z) = 2\pi i \times \frac{\pi i}{\exp(a\pi i)}$$

Observe that in the limit

$$\int_{\Gamma_0} f(z) \; dz = \int_0^\infty \frac{\ln x}{x^a (1+x)} \; dx = J.$$

For $\Gamma_1$ we have by ML estimate $\lim_{R\to\infty} 2\pi R \times (\log R + 2\pi) / R^a / (R-1) $ or $\lim_{R\to\infty} 2\pi \times (\log R + 2\pi) / R^a + \lim_{R\to\infty} 2\pi \times (\log R + 2\pi) / R^a /(R-1) = 0,$ so this vanishes.

Furthermore for $\Gamma_2$ in the limit (we get $\mathrm{Log}(z) = \log |z| + 2\pi i$)

$$\int_{\Gamma_2} f(z) \; dz = \int_\infty^0 \frac{\ln x + 2\pi i} {\exp(2a\pi i) x^a(1+x)} \; dx \\ = - \exp(-2a\pi i) J - \exp(-2a\pi i) 2\pi i \int_0^\infty \frac{1}{x^a(1+x)} \; dx \\ = - \exp(-2a\pi i) J - \exp(-2a\pi i) 2\pi i K.$$

For $\Gamma_3$ we again use ML and get $\lim_{\epsilon\to 0} 2 \pi \epsilon \times (|\log\epsilon| +2\pi) / \epsilon^a / (1-\epsilon)$ which is $\lim_{\epsilon\to 0} 3/2 \times 2 \pi \epsilon^{1-a} \times (|\log\epsilon| +2\pi) = 0$, so this too vanishes. We have shown that in the limit

$$2\pi i \times \frac{\pi i}{\exp(a\pi i)} = J - \exp(-2a\pi i) J - \exp(-2a\pi i) 2\pi i K.$$

This is

$$-2\pi^2 = 2 i J \sin(a\pi) - \exp(-a\pi i) 2\pi i K.$$

It seems we require $K.$ We use the same estimates and the same contour to get

$$K (1-\exp(-2a\pi i)) = 2\pi i \times \exp(-a\pi i)$$

which is

$$K = 2\pi i \frac{\exp(-a\pi i)}{1-\exp(-2a\pi i)} = 2\pi i \frac{1}{\exp(a\pi i)-\exp(-a\pi i)} = \frac{\pi}{\sin(a\pi)}.$$

We then get

$$2i J\sin(a\pi) = -2\pi^2 + \exp(-a\pi i)2\pi i \frac{\pi}{\sin(a\pi)} \\ = 2\pi^2 \frac{-\sin(a\pi)+i\exp(-a\pi i)}{\sin(a\pi)} \\ = 2\pi^2 \frac{-\exp(a\pi i)/2/i+\exp(-a\pi i)/2/i-\exp(-a\pi i)/i} {\sin(a\pi)} \\ = - 2\pi^2 \frac{\cos(a\pi)/i}{\sin(a\pi)}.$$

We thus have

$$J = - 2\pi^2 \frac{\cos(a\pi)/i}{\sin^2(a\pi)} \frac{1}{2i}$$

This is

$$\bbox[5px,border:2px solid #00A000]{ J = \pi^2 \frac{\cos(a\pi)}{\sin^2(a\pi)}.}$$