# Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$

While I was working on this question by @Vladimir Reshetnikov, I've conjectured the following closed-forms. $$I_0(n)=\int_0^\infty \frac{1}{\left(\cosh x\right)^{1/n}} \, dx \stackrel{?}{=} \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\tfrac{1}{2n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{2n}\right)},$$ for all $n\geq1$ real numbers. In another form:

$${_2F_1}\left(\begin{array}c\tfrac{1}{2n},\tfrac1n\\1+\tfrac{1}{2n}\end{array}\middle|\,-1\right) \stackrel{?}{=} \frac{\sqrt{\pi}}{n\,2^{1+\frac1n}} \frac{\Gamma\left(\tfrac{1}{2n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{2n}\right)}.$$ Another conjectured closed-form is $$I_1(n)=\int_0^\infty \frac{1}{\left(1+\cosh x\right)^{1/n}} \, dx \stackrel{?}{=} \frac{\sqrt{\pi}}{2^{1/n}} \frac{\Gamma\left(\tfrac{1}{n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{n}\right)},$$ for all $n \geq 1$ real numbers. In another form: $${_2F_1}\left(\begin{array}c\tfrac1n,\tfrac2n\\1+\tfrac{1}{n}\end{array}\middle|\,-1\right) \stackrel{?}{=} \frac{\sqrt{\pi}}{n\,2^{\frac2n}} \frac{\Gamma\left(\tfrac{1}{n}\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{n}\right)}.$$

Here $\cosh$ is the hyperbolic cosine function, $\Gamma$ is the gamma function, and ${_2F_1}$ is the hypergeometric function.

Questions.

• $1^{\text{st}}$ question. How could we prove the conjectured closed-form for $I_0$ and $I_1$.
• $2^{\text{nd}}$ question. How could we show the equivalent hypergeometric forms.
• $3^{\text{rd}}$ question. There is a closed-form of $I_a(n) = \int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a\geq0,n\geq1$ real numbers in term of Appell $F_1$ function. Could we get a closed-form just in term of gamma function?

For the first one, \begin{align} \int_0^{\infty} (\operatorname{sech}x)^{2s}dx \\ &= \int_0^{\infty} (\operatorname{sech}^2x)^{s-1}\operatorname{sech}^2x\, dx \\ &= \int_0^{\infty} (1-\tanh^2x)^{s-1}\,\mathrm{d}(\tanh x)\\ &= \int_0^1 (1-x^2)^{s-1} \mathrm{d}x\\ &= \frac12 \int_0^1 (1-x)^{s-1} x^{-\frac12} \mathrm{d}x\\ &= \frac12 B(s,\frac12)=\frac{\sqrt{\pi}}{2}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align} and so your conjecture is correct. For the second integral, let $x=2t$: \begin{align} \int_0^{\infty} \frac{1}{(1+\cosh x)^s}\mathrm{d}x \\ &= 2\int_0^{\infty} \frac{1}{(1+\cosh 2t)^s}\mathrm{d}t \\ &= 2\int_0^{\infty} \frac{1}{(2\cosh^2(t))^s}\mathrm{d}t\\ &= 2^{1-s} \int_0^{\infty} (\operatorname{sech}t)^{2s}\mathrm{d}t\\ &= 2^{-s} B(s,\frac12)=\frac{\sqrt{\pi}}{2^s}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align}

• Very well done! +1 – Mark Viola Jul 30 '15 at 20:28
• Simple and nice trick! (+1) – Sangchul Lee Jul 30 '15 at 21:29
• Beat me by four hours ! :-$)$ – Lucian Jul 30 '15 at 23:53

Extending @nospoon's idea, we notice that

$$a + \cosh 2x = (a+1)\cosh^2 x (1 - b \tanh^2 x), \qquad b =\frac{a-1}{a+1}.$$

If $a > -1$, then $b < 1$ and using the substitution $u = \tanh^2 x$ we get

$$\int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{(1 - u)^{s-1}}{(1 - b u)^s\sqrt{u}} \, du.$$

Making further substitution $v = \frac{1-u}{1-bu}$, we have

$$\int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv.$$

This is easily integrated when $b = 0$ or $b = -1$, each correspondingto $a = 1$ or $a = 0$, but I doubt that this integral has a nice closed form in general. For example, when $s = 1/2$, this becomes an elliptic integral and Mathematica 11 gives

$$\int_{0}^{1} \frac{dv}{\sqrt{v(1-v)(1-bv)}} = 2 \left[ \frac{1}{\sqrt{b}} K \left( \sqrt{\tfrac{1}{b}} \right) + i K \left( \sqrt{1-b} \right) \right], \quad 0 < b < 1,$$

where $K(k)$ is the complete elliptic integral of the 1st kind.

• Thank you for your solution. +1. Is there a closed-form in term of $K$ function for $s=1/4$? – user153012 Jul 30 '15 at 21:45
• Can you check your last formula? Doesn't work for $b=1/4$ (RHS is not real) – Wolfgang Mar 3 '18 at 15:45
• @Wolfgang Thank you for pointing out an error. I fixed it. – Sangchul Lee Mar 3 '18 at 17:49
• well, not yet: wolframalpha.com/input/… – Wolfgang Mar 4 '18 at 19:29
• @Wolfgang Notice in the comment of the WolframAlpha page that the convention $K(m)$ is used where $m=k^2$, but in my answer the convention $K(k)$ is used. To correctly apply the formula in WolframAlpha, you have to strip off inner squre roots. – Sangchul Lee Mar 4 '18 at 19:32

Although not as simple as nospoon's answer, here's another approach for the first one.

For $s>0$,

\begin{align} \int_{0}^{\infty} \frac{1}{\cosh^{s}(x)} \, dx &= 2^{s} \int_{0}^{\infty} \frac{1}{(e^{x}+e^{-x})^{s}} \\ &= 2^{s} \int_{0}^{1} \frac{1}{(u^{-1}+u)^{s}} \, \frac{du}{u} \\ &= 2^{s} \int_{0}^{1} \frac{u^{s-1}}{(u^{2}+1)^{s}} \, du \\ &= 2^{s} \int_{0}^{1} \frac{(\sqrt{v})^{s-1}}{(v+1)^{s}} \, \frac{dv}{2 \sqrt{v}} \\ &=2^{s-1} \int_{0}^{1} \frac{v^{s/2-1}}{(v+1)^{s}} \, dv \\ &=2^{s-1} \, B \left(\frac{s}{2}, 1 \right) \, _2F_1 \left(s,\frac{s}{2}; 1+ \frac{s}{2}; -1 \right) \tag{1}\\ &=2^{s-1} \, B \left(\frac{s}{2}, 1 \right) \frac{\Gamma \left(1+ \frac{s}{2} \right) \Gamma \left(1+ \frac{s}{2} \right)}{\Gamma \left(1+s \right) \Gamma\left(1 \right)} \tag{2} \\ &= 2^{s-1} \, \Gamma \left(\frac{s}{2} \right) \frac{\Gamma \left(1+ \frac{s}{2} \right)}{\Gamma(1+s)} \\ &= \frac{\sqrt{\pi}}{2} \frac{ \Gamma (\frac{s}{2})}{\Gamma(\frac{1}{2}+ \frac{s}{2})} \tag{3} \end{align}

$(2)$ Kummer's theorem