Evaluate $\int_{0}^{\infty} \frac{dx}{1+x^\lambda} = \frac{\frac{\pi}{\lambda}}{\sin\left(\frac{\pi}{\lambda}\right)} $ for $\lambda \geq 2$

Question

\begin{align} \int_{0}^{\infty} \frac{dx}{1+x^\lambda} = \frac{\frac{\pi}{\lambda}}{\sin\left(\frac{\pi}{\lambda}\right)} \end{align} where $\lambda \geq 2$. First I know it converges by $p$ test.

For even case, I know how to do this integral. Simply take the upper half-plane and do contour integral.

\begin{align} \int_0^{\infty} \frac{dx}{1+x^{2m}} &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{dx}{1+x^{2m}} = \frac{1}{2} 2\pi i \sum_{k=0}^{m-1} \operatorname{Res}[f;z_k] \\ &= - \frac{\pi i}{m} \frac{e^{i \frac{\pi}{2m}}}{1-e^{i\frac{\pi}{m}}} = \frac{\frac{\pi}{2m}}{\sin \left(\frac{\pi}{2m}\right)} \end{align} Note $z_0, \cdots, z_{m-1}$ are simple poles which located inside upper half-plane. Actually this is what Cauchy do in his paper(1814). i.e., he compute following integral for $n>m$ \begin{align} \int_{-\infty}^{\infty} \frac{x^{2m}}{1+x^{2n}}dx = \frac{\pi}{n \sin\left(\frac{2m+1}{2n} \pi \right)} \end{align}

Now I want to evaluate the integral for odd cases also. How about odd cases? Is there a nice way to evaluate this integral? or is there any simple way to compute the general case?

Answer 1

In this answer, let $\lambda$ be a complex number such that $\text{Re}(\lambda)>1$. Let $U_\lambda$ be the set of complex numbers that lie between the line $\text{Im}(z)=0$ and the line $\text{Im}(z)=-\dfrac{2\pi\,\text{Im}(\lambda)}{|\lambda|^2}$. Define $$f(z):=\frac{\exp(z)}{1+\exp(\lambda z)}$$ for each $z\in \mathbb{C}\setminus\dfrac{2\pi\text{i}}{\lambda}\left(\mathbb{Z}+\dfrac{1}{2}\right)$. Denote by $I(\lambda)$ the integral $$I(\lambda):=\int_0^\infty\,\frac{1}{1+x^\lambda}\,\text{d}x=\int_{-\infty}^{+\infty}\,\frac{\exp(t)}{1+\exp(\lambda t)}\,\text{d}t\,.$$ We shall prove that $$I(\lambda)=\frac{\pi}{\lambda}\,\text{csc}\left(\frac{\pi}{\lambda}\right)\,.$$ For $R>0$, let $\mathscr{C}(R)$ be the contour given by $$\begin{align} \left[-R,+R\right]&\cup \left[+R,+R+\frac{2\pi\text{i}}{\lambda}\right]\cup\left[+R+\frac{2\pi\text{i}}{\lambda},-R+\frac{2\pi\text{i}}{\lambda}\right]\cup\left[-R+\frac{2\pi\text{i}}{\lambda},-R\right]\,.\end{align}$$ It is easily seen that $$\lim_{R\to \infty} \,\int_{\mathscr{C}(R)}\,f(z)\,\text{d}z=\Biggl(1-\exp\left(\frac{2\pi\text{i}}{\lambda}\right)\Biggr)\,I(\lambda)\,.$$ On the other hand, for every $R>0$, $\mathscr{C}(R)$ encloses exactly one pole of $f$, which is $\dfrac{\pi\text{i}}{\lambda}$. (All poles of $f(z)$ are $z=\dfrac{(2n+1)\pi\text{i}}{\lambda}$, where $n\in\mathbb{Z}$, and only $z=\dfrac{\pi\text{i}}{\lambda}$ lies in $U_\lambda$.) Consequently, $$\begin{align}\int_{\mathscr{C}(R)}\,f(z)\,\text{d}z&={2\pi\text{i}}\,\text{Res}_{z=\frac{\pi\text{i}}{\lambda}}\big(f(z)\big) ={2\pi\text{i}}\,\lim_{z\to\frac{\pi\text{i}}{\lambda}} \frac{\exp(z)}{\lambda\,\exp(\lambda z)} \\&={2\pi\text{i}}\,\frac{\exp\left(\frac{\pi\text{i}}{\lambda}\right)}{\lambda \,\exp(\pi\text{i})}=-\frac{2\pi\text{i}}{\lambda}\,\exp\left(\frac{\pi\text{i}}{\lambda}\right)\,.\end{align}$$ Therefore, $$\begin{align}I(\lambda)&=\frac{\lim\limits_{R\to \infty} \,\int_{\mathscr{C}(R)}\,f(z)\,\text{d}z}{1-\exp\left(\frac{2\pi\text{i}}{\lambda}\right)} =\frac{-\frac{2\pi\text{i}}{\lambda}\,\exp\left(\frac{\pi\text{i}}{\lambda}\right)}{1-\exp\left(\frac{2\pi\text{i}}{\lambda}\right)} \\&=\frac{\pi}{\lambda}\,\left(\frac{\exp\left(+\frac{\pi\text{i}}{\lambda}\right)-\exp\left(-\frac{\pi\text{i}}{\lambda}\right)}{2\text{i}}\right)^{-1}=\frac{\pi}{\lambda}\,\text{csc}\left(\frac{\pi}{\lambda}\right)\,.\end{align}$$

Answer 2

Use $t=x^\lambda$ then $\mathrm dx=\frac1\lambda t^{\frac1\lambda-1} \,\mathrm dt$ so that $$\int_{0}^{\infty} \frac{\mathrm dx}{1+x^\lambda}=\frac1\lambda\int_0^\infty t^{\frac1\lambda-1} (1+t)^{-1}\,\mathrm dt=\frac1\lambda\operatorname B\left(\frac1\lambda,1- \frac1\lambda\right),$$

where I have used the fact that the Beta function satisfies

$$\operatorname{B}(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,\mathrm dt$$ for $x,y>0$.

Now, $$\operatorname{B}\left(\frac1\lambda,1- \frac1\lambda\right)=\frac{\Gamma(1/\lambda)\Gamma(1-1/\lambda)}{\Gamma(1)}=\Gamma(1/\lambda)\Gamma(1-1/\lambda).$$

By Euler's reflection formula, it follows that $$\Gamma(1/\lambda)\Gamma(1-1/\lambda)=\frac{\pi}{\sin(\pi/\lambda)}$$ and we are done.

Remark. This works for all real $\lambda>1$.

Answer 3

Actually, the hard part about proving the reflection formula for the Gamma function is to establish that $$\int_0^{\infty}\frac{t^{y-1}}{1+t}dt=\frac{\pi}{\sin\pi y}$$ for $0<y<1$. But just proving the theorem you have for even powers, i.e. that $$\frac{\pi}{2n\sin\left(\frac{2m+1}{2n}\pi\right)}=\int_0^{\infty}\frac{x^{2m}}{1+x^{2n}}dx=\frac1{2n}\int_0^{\infty}\frac{x^{\frac{2m+1}{2n}-1}}{1+x}dx$$ Shows that the theorem is valid for rational $0<y<1$ with odd numerators and even denominators. We can create a sequence $\{y_k\}$ by, for example, truncating the decimal representation of any $y$ after $k$ digits and adding $10^{-k}$ if the $k^{\text{th}}$ digit were even. Then $\lim_{k\rightarrow\infty}y_k=y$ and assuming we know somehow that the above integral is a continuous function of $y$ then we can see that the above theorem is true.

Then you can find $$\int_0^{\infty}\frac{x^p}{1+x^{2q+1}}dx=\frac1{2p+1}\int_0^{\infty}\frac{x^{\frac{p+1}{2q+1}-1}}{1+x}dx=\frac{\pi}{(2q+1)\sin\left(\frac{p+1}{2q+1}\pi\right)}$$ For any $1\le p\le2q$. But the easy way to prove the theorem is via a keyhole contour and I know there is a worked example somewhere in this forum but Im just not good enough at searching to find it.

EDIT: I added an $\epsilon$-$\delta$ proof of continuity of $f(y)=\int_0^{\infty}\frac{t^{y-1}}{t+1}dt$ in another post where I attempted to prove that $\int_0^{\infty}\frac{t^{y-1}}{1+t}dt=\frac{\pi}{\sin\pi y}$ by elementary means, so the continuity required to extend the theorem from decimal fractions to all $y\in(0,1)$ is valid.