integration - Calculating the integral $int_0^{infty}{frac{ln x}{1+x^n}}$ using complex analysis

I need to calculate $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ $,n\geq2$ using complex analysis. I probably need to use the Residue Theorem.

I use the function $f(z)={\frac{\ln z}{1+z^n}}$ in the normal branch.

I've tried to use this contour.

Where $\theta$ is an angle so that only $z_{0}$ will be in the domain (I hope this is clear from the drawing)

I estimated $|\int_{\Gamma_{R}}|$ and $|\int_{\Gamma_{\epsilon}}|$ and showed that they must tend to $0$ when $\epsilon \rightarrow0$ and $R \rightarrow\infty$. (Is it true?)

However I'm having trouble calculating $\int_{\Gamma_{2}}$ . Does it have something to do with choosing the "right" $\theta$?

Any ideas?

Thanks!

UPDATE

After Christopher's comment I chose $\theta=\frac{2\pi}{n}$ which gives, after the paramatrization $\Gamma (t) = te^{\frac{2\pi i}{n}}$, $t\in(\epsilon,R)$:

$$\int_{\Gamma_{2}}{\frac{\ln z}{1+z^n}dz} = \int_{\epsilon}^{R}{\frac{\ln (te^\frac{2\pi i}{n})}{1+t^n}e^\frac{2\pi i}{n}dt} = e^\frac{2\pi i}{n}\int_{\epsilon}^{R}{\frac{\ln t}{1+t^n}dt} + ie^\frac{2\pi i}{n}\int_{\epsilon}^{R}{\frac{\frac{2\pi}{n}}{1+t^n}dt} =$$ $$= e^\frac{2\pi i}{n}\int_{0}^{\infty}{\frac{\ln t}{1+t^n}dt} + ie^\frac{2\pi i}{n}\int_{0}^{\infty}{\frac{\frac{2\pi}{n}}{1+t^n}dt}$$

But I have no idea how to deal with the second integral.

Answer

Let us define, for $r>0$, $$\Gamma_{r}=\{r e^{i\alpha}\ :\ 0\leq\alpha\leq\theta\}$$ The circular arcs of your contour are then $\Gamma_{\epsilon}$ and $\Gamma_{R}$.

We note that, for $z\in\Gamma_{\epsilon}$ $$\left|\frac{\ln z}{1+z^n}\right|\leq|\ln z|\leq 2\ln\epsilon$$ if $\epsilon$ is small enough. So $$\left|\int_{\Gamma_\epsilon} f(z)\right|\leq 2\theta\epsilon\ln\epsilon$$ which goes to $0$ with $\epsilon$. Also, for $R$ large enough, $$|\ln z|\leq 2\ln R$$ when $z\in \Gamma_R$. So $$\left|\frac{\ln z}{1+z^n}\right|\leq\frac{2\ln R}{R^n-1}$$ hence $$\left|\int_{\Gamma_R} f(z)\right|\leq\theta R\frac{2\ln R}{R^n-1}$$ which goes to $0$ when $R\to\infty$, if $n>1$. (So, yes, these two pieces go to zero)

Now take $\Gamma_1=\{r\ :\ \epsilon\leq r\leq R\}$ and $\Gamma_2=e^{i\theta}\Gamma_1$; define the contour $\Gamma$ by walking along $\Gamma_1$, then $\Gamma_R$, then backwards along $\Gamma_2$ and $\Gamma_\epsilon$.

The roots of $1+z^n=0$ are the $n$-th roots of $-1$, that is $$z_k=e^{i\frac{(2k+1)\pi}{n}}$$ so, if we take $\theta\in (\frac{\pi}{n},\frac{3\pi}{n})$, $\Gamma$ surrounds exactly one pole of $f$.

Now, following the hint of Christopher A. Wong, we set $\theta=\frac{2\pi}{n}$; therefore, if $z\in\Gamma_2$, we get $z^n=|z|^ne^{in\theta}=|z|^n$ and $\ln z=\ln|z| + i\frac{2\pi}{n}$.

So $$\int_{\Gamma_2}f(z)=\int_{\epsilon}^R\frac{\ln x + i\frac{2\pi}{n}}{1+x^n}e^{i\frac{2\pi}{n}}dx$$

We recall that $$\int_0^\infty\frac{1}{1+x^n}dx=\frac{\pi/n}{\sin(\pi/n)}$$ with the same method we used now (integrating on $\Gamma$!).

Call $J=\int_{0}^\infty \ln x/(1+x^n) dx$ and $L$ the residue of $f(z)=\ln z /(1+z^n)$ at $z=e^{i\frac{\pi}{n}}$, then $$2i\pi L=J-e^{i\frac{2\pi}{n}}\left(J+i\frac{2\pi}{n}\frac{\pi/n}{\sin(\pi/n)}\right)$$ that is $$J=\frac{2i\pi L+ie^{i\frac{2\pi}{n}}\frac{2\pi^2/n^2}{\sin(\pi/n)}}{1-e^{i\frac{2\pi}{n}}}$$

Now it is enough to perform the computations.