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

\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?

• You can use contour integral with the contour formed by the sector of the circle of radius $r$ centered at the origin which is supported by the arc from $r$ to $re^{2\pi i/\lambda}$. For $r>1$, the function $\frac{1}{1+z^\lambda}$ has a unique pole inside the contour at $z=e^{i\pi/\lambda}$. Mar 9, 2020 at 14:26

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 $$\lambda>1$$.

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}

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 $$0. But just proving the theorem you have for even powers, i.e. that $$\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 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.