Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$?

Question

As mentioned in my post, I started to investigate the integral $$ I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}, $$ where $m$ and $n$ are natural number and $a$ is positive.

First of all, let’s start with the ‘simple’ case, $$ I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}. $$

However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula

$$ \int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right). $$

which I can’t prove by elementary method yet. Please tell me if you have any.

Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]} =\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$

Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields $$ \begin{aligned} \int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right) \cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}} \end{aligned} $$

Rearranging and simplifying yields $$ \boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)} $$

Putting $a=1$ gives our formula $$ \boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)} $$

For verification, $$ \begin{aligned} \int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26 =\frac{1118260 \pi}{4782969 \sqrt{3}}, \end{aligned} $$ which is checked by Wolframalpha $$\begin{aligned} \int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\ &=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\ &\doteq 0.845906950943631, \end{aligned} $$ which is checked by Wolframalpha

My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$

Answer 1

Without using the formula $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),$ I thought of the Gamma Function. $$ \begin{aligned} \text { Let } t &=\frac{1}{x^{n}+1}, \text { then } x=\left(\frac{1}{t}-1\right)^{\frac{1}{n}} \\ d x &=\frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1-n}{n}}\left(-\frac{1}{t^{2}} d t\right)=-\frac{(1-t)^{\frac{1}{n}-1}}{n t ^{\frac{1+n}{n }} }d t \end{aligned} $$

The integral is transformed into $$ \begin{aligned} \int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}&=\frac{1}{n} \int_{0}^{1} t^{m} \cdot \frac{(1-t)^{\frac{1-n}{n}}}{t^{\frac{1+n}{n}}} d t \\ &=\frac{1}{n} \int_{0}^{1} t^{\frac{m n-1}{n}-1}(1-t)^{\frac{1}{n} -1}dx \\ &=\frac{1}{n} B\left(m -\frac{1}{n}, \frac{1}{n}\right) \end{aligned} $$

Now applying the formula $$ B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}, $$

we can now conclude that

$$\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}= \frac{\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)}{n(m-1)!}}$$

Letting $x\mapsto\frac{x}{\sqrt[n]a}$ yields the general integral,

$$\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}= \frac{\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)}{n a^{m-\frac{1}{n}}(m-1)!}}$$

Taking the same integral for example, we have $$ \int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}= \frac{\Gamma\left(\frac{1}{12}\right) \Gamma\left(\frac{59}{12}\right)}{12 \Gamma(5)}= \frac{1}{288} \Gamma\left(\frac{1}{12}\right) \Gamma\left(\frac{59}{12}\right)\approx 0.845907, $$ which is checked by Wolframalpha.

However, I don’t know how to further simplify the formula $$\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right),$$

can you help?

$$\textrm{ ********} \tag*{} $$

Thanks to Mr Clathratus who gave me a nice simplification to $$\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right).$$ Using the identity $$ \Gamma(x+1)=x \Gamma(x), $$ we have $$ \begin{aligned} \Gamma\left(m-\frac{1}{n}\right) &=\left(m-1-\frac{1}{n}\right) \Gamma\left(m-\left(-\frac{1}{n}\right)\right.\\ &=\left(m-1-\frac{1}{n}\right)\left(m-1-\frac{1}{n}\right) \Gamma\left(m-2-\frac{1}{n}\right) \\ & \qquad\qquad \vdots \\ &=\left(m-1-\frac{1}{n}\right)\left(m-2-\frac{1}{n}\right) \cdots\left(1-\frac{1}{n} \right)\Gamma\left(1-\frac{1}{n}\right)\\ &= \Gamma\left(1-\frac{1}{n}\right) \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) \\ \Gamma\left(m-\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right) &=\prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) \Gamma\left(1-\frac{1}{n} \right)\Gamma\left(\frac{1}{n}\right) \end{aligned} $$ By the Euler's Reflection Theorem, $$ \Gamma\left(1-\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)=\pi \csc \left(\frac{\pi}{n}\right) $$ Putting back, we now obtain the same formula as before $$ \boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! na^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k-\frac{1}{n} )} $$

Answer 2

Proving $\displaystyle\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$:

Set $a=z$ and $b=1-z$ in the Beta function: $$\operatorname{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}\mathrm{d}x,$$ we obtain \begin{gather} \Gamma(z)\Gamma(1-z)=\int_0^\infty\frac{x^{z-1}}{1+x}\mathrm{d}x\\ =\left(\int_0^1+\int_1^\infty\right) \frac{x^{z-1}}{1+x}\mathrm{d}x=\int_0^1 \frac{x^{z-1}}{1+x}\mathrm{d}x+\underbrace{\int_1^\infty \frac{x^{z-1}}{1+x}\mathrm{d}x}_{x\to 1/x}\\ =\int_0^1 \frac{x^{z-1}}{1+x}\mathrm{d}x+\int_0^1 \frac{x^{-z}}{1+x}\mathrm{d}x\\ \left\{\text{expand $\frac1{1+x}$ in series in both integrals}\right\}\\ =\int_0^1 x^{z-1}\left(\sum_{n=0}^\infty(-1)^nx^n\right)\mathrm{d}x+\int_0^1 x^{-z}\left(\sum_{n=0}^\infty(-1)^nx^n\right)\mathrm{d}x\\ \{\text{interchange integration and summation}\}\\ =\sum_{n=0}^\infty(-1)^n\int_0^1 x^{n+z-1}\mathrm{d}x+\sum_{n=0}^\infty(-1)^n\int_0^1 x^{n-z}\mathrm{d}x\\ =\sum_{n=0}^\infty\frac{(-1)^n}{n+z}+\sum_{n=0}^\infty\frac{(-1)^n}{n-z+1}\\ \{\text{seperate the first term of the first sum and shift the index of the second}\}\\ =\frac1z+\sum_{n=1}^\infty\frac{(-1)^n}{n+z}-\sum_{n=1}^\infty\frac{(-1)^n}{n-z}\\ =\frac1z-\sum_{n=1}^\infty\frac{2z(-1)^n}{n^2-z^2}\\ =\frac{\pi}{\sin(\pi z)}. \end{gather}

To prove the last equality, set $x=0$ in the Fourier series of $\cos(zx)$: \begin{equation} \cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}-\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{n^2-z^2}\right],\quad z\notin\mathbb{Z}.\label{cos(zx)} \end{equation}

Thus,

$$\Gamma(z)\Gamma(1-z)=\int_0^\infty\frac{x^{z-1}}{1+x}\mathrm{d}x=\frac{\pi}{\sin(\pi z)}.$$

To complete the proof, replace $z$ by $1/n$ then let $x=y^n$.

Answer 3

Glad to see that there are several nice solutions to the problem, now I want to add one more by discovering a reduction formula using integration by parts.

In order to use integration by parts, I applied the inversion substitution to the original integral. For any fixed natural number $n$, we now define

$$ J_{m}:=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} \stackrel{x \rightarrow \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{m n-2}}{\left(x^{n}+1\right)^{m}} d x $$ Noting that $$ \frac{d}{d x}\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)=-(m-1) \cdot \frac{nx^{n-1}}{\left(x^{n}+1\right)^{m}}, $$ we have $$J_m=-\frac{1}{n(m-1)} \int_{0}^{\infty} x^{m n-n-1} d\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right) $$ By integration by parts, we get $$ \begin{aligned} J_{m} &\left.=-\frac{m-1}{n}\left[\frac{x^{m n-n-1}}{\left(x^{n}+1\right)^{m-1}}\right]_{0}^{\infty}+\frac{1}{n(m-1)} \int_{0}^{\infty} \frac{(m n-n-1) x^{m n-1-2}}{\left(x^{n}+1\right)^{m-1}}\right] \\ &=\frac{m n-n-1}{n(m-1)} \int_{0}^{\infty} \frac{x^{m n-n+2}}{\left(x^{n}+1\right)^{m-1}} d x\\&= \frac{\left(m-1-\frac{1}{n}\right)}{m-1} J_{m-1} \end{aligned} $$ Applying the reduction formula $m$ times yields inductively \begin{aligned} J_m&=\frac{m-1-\frac{1}{n}}{m-1} \cdot \frac{m-2-\frac{1}{n}}{m-2} \cdot \frac{m-3-\frac{1}{n}}{m-3} \cdot \frac{1-\frac{1}{n}}{1} J_{0} \\ & = \frac{1}{(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) J_{0} \end{aligned}

Using the well-known formula, $$ J_0=\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right), $$ we can conclude that $$\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} =\frac{\pi \csc \left(\frac{\pi}{n}\right)}{n(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) $$

Answer 4

You could make it faster using $$I=\int \frac{d x}{\left(x^{n}+a\right)^{m}}=x\, a^{-m} \, _2F_1\left(m,\frac{1}{n};1+\frac{1}{n};-\frac{x^n}{a}\right)$$ which gives $$J=\int_0^\infty \frac{d x}{\left(x^{n}+a\right)^{m}}=a^{\frac{1}{n}-m}\,\frac{\Gamma \left(1+\frac{1}{n}\right) \Gamma \left(m-\frac{1}{n}\right)}{\Gamma (m)}$$ provided $\Re(m n)>1\land \Re(n)>0$.