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Inspired by my post, I start to investigate a more general integral

$$ J_{m}=\int_{0}^{\infty} \frac{1}{\left(x^{m}+\frac{1}{x^{m}}\right)^{2}} d x \quad \text { where } m \in \mathbb{N} . \\$$ Rearranging and integrate by parts yields

\begin{align*} J_{m} &=\int_{0}^{\infty} \frac{x^{2 m}}{\left(x^{2 m}+1\right)^{2}} d x \\ &=-\frac{1}{2 m} \int_{0}^{\infty} x d\left(\frac{1}{x^{2 m}+1}\right) \\ &=-\left[\frac{x}{2 m\left(x^{2 m}+1\right)}\right]_{0}^{\infty}+\frac{1}{2 m} \int_{0}^{\infty} \frac{1}{x^{2 m}+1} d x \\ &=\frac{1}{2 m} \cdot\frac{\pi}{2 m} \csc \left(\frac{\pi}{2 m}\right) \tag{$*$}\\ &=\frac{\pi}{4 m^{2}} \csc \left(\frac{\pi}{2 m}\right) \end{align*} where $(*)$ comes from the Theorem, $$ \int_{0}^{\infty} \frac{d x}{1+x^{n}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right). $$

In particular, $$ \begin{array}{l} \displaystyle I_{3}=\frac{\pi}{36} \csc \left(\frac{\pi}{6}\right)=\frac{\pi}{18} \\ \displaystyle I_{4}=\frac{\pi}{64} \csc \left(\frac{\pi}{8}\right)=\frac{\pi}{32 \sqrt{2-\sqrt{2}}} \end{array} $$

My question:

Can we go further for $$ I(m, n):=\int_{0}^{\infty} \frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}. $$

where $m,n\in \mathbb{N}$ and $n\geq 2$?

Latest edit

I have found an answer for $I(m,n)$ using the result in one of my posts and share with you now. $$ I(m,n)=\frac{\pi}{2m(n-1) !} \csc \frac{(mn+1) \pi}{2m} \prod_{j=1}^{n-1}\left(j-\frac{mn+1}{2m}\right),\tag*{} $$

However, the proof is a bit complicated. Can you help give a simpler one?

Gary
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Lai
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3 Answers3

8

Too long for a comment

We can consider a general case: $$I(a,b,c)=\int_0^\infty\frac{x^a}{(x^b+1)^c}dx$$ Making the substitution $x^b=t$ $$I(a,b,c)=\frac{1}{b}\int_0^\infty\frac{t^{\frac{a+1}{b}-1}}{(t+1)^c}dt$$ Making another substitution $x=\frac{1}{1+t}$ $$I(a,b,c)=\frac{1}{b}\int_0^1x^{c-\frac{a+1}{b}-1}(1-x)^{\frac{a+1}{b}-1}=\frac{1}{b}\frac{\Gamma\big(c-\frac{a+1}{b}\big)\Gamma\big(\frac{a+1}{b}\big)}{\Gamma(c)}$$ For our specific case $a=mn, b=2m, c=n$, and we get $$I(m, n)=\frac{\Gamma\big(\frac{n}{2}-\frac{1}{2m}\big)\Gamma\big(\frac{n}{2}+\frac{1}{2m}\big)}{2m\,\Gamma(n)}$$

Svyatoslav
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2

Still not elementary but quite fast.

$$I_{m,n}=\int\frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}=-\frac{ \, _2F_1\left(n,\frac{m n-1}{2 m};\frac{m (n+2)-1}{2 m};-x^{-2 m}\right)}{(m n-1)\,x^{mn-1}}$$ $$J_{m,n}=\int_0^\infty\frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}=\frac{\Gamma \left(\frac{m n+1}{2 m}\right) \Gamma \left(\frac{m (n+2)-1}{2 m}\right)}{(m n-1) \Gamma (n)}$$

Claude Leibovici
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  • Nice(+1). Given that $$\Gamma \Big(\frac{m (n+2)-1}{2m}\Big)=\Gamma \Big(\frac{n}{2}-\frac{1}{2m}+1\Big)=\Gamma \Big(\frac{n}{2}-\frac{1}{2m}\Big)\frac{mn-1}{2m}$$ our answers coincide – Svyatoslav Feb 06 '22 at 10:35
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    @Svyatoslav. I totally agree. In fact, beside this post, we had sveral similar answers. Cheers :-) and (+1) – Claude Leibovici Feb 06 '22 at 10:37
1

I first convert the integral $$I(m,n)=\int_{0}^{\infty} \frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}} $$into

$$ I(m, n)=\int_{0}^{\infty} \frac{x^{m n}}{\left(x^{2 m}+1\right)^{n}} d x . $$

Using the result in my post,

$$ \int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+1\right)^{n}}=\frac{\pi}{m(n-1) !} \csc \frac{(r+1) \pi}{m} \prod_{j=1}^{n-1}\left(j-\frac{r+1}{m}\right),\tag*{} $$

we can conclude that $$ I(m,n)=\int_{0}^{\infty} \frac{x^{mn} d x}{\left(x^{2m}+1\right)^{n}}=\frac{\pi}{2m(n-1) !} \csc \frac{(mn+1) \pi}{2m} \prod_{j=1}^{n-1}\left(j-\frac{mn+1}{2m}\right),\tag*{} $$

However, the solution is a bit complicated and non-elementary.

Is there any simpler or more elementary method?

Lai
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