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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 can we prove the conjectured closed form for $I_0$ and $I_1$?
  • $2^{\text{nd}}$ question. How can 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 the gamma function?
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3 Answers 3

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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 $x=2t$: $$\begin{align} \int_0^{\infty} \frac{1}{(1+\cosh x)^s}\mathrm{d}x \\ &= 2\int_0^{\infty} \frac{1}{(1+\cosh 2t)^s}\mathrm{d}t \\ &= 2\int_0^{\infty} \frac{1}{(2\cosh^2(t))^s}\mathrm{d}t\\ &= 2^{1-s} \int_0^{\infty} (\operatorname{sech}t)^{2s}\mathrm{d}t\\ &= 2^{-s} B(s,\frac12)=\frac{\sqrt{\pi}}{2^s}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align}$$

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    $\begingroup$ Very well done! +1 $\endgroup$
    – Mark Viola
    Jul 30, 2015 at 20:28
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    $\begingroup$ Simple and nice trick! (+1) $\endgroup$ Jul 30, 2015 at 21:29
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    $\begingroup$ Beat me by four hours ! :-$)$ $\endgroup$
    – Lucian
    Jul 30, 2015 at 23:53
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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.

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  • $\begingroup$ Thank you for your solution. +1. Is there a closed-form in term of $K$ function for $s=1/4$? $\endgroup$
    – user153012
    Jul 30, 2015 at 21:45
  • $\begingroup$ Can you check your last formula? Doesn't work for $b=1/4$ (RHS is not real) $\endgroup$
    – Wolfgang
    Mar 3, 2018 at 15:45
  • $\begingroup$ @Wolfgang Thank you for pointing out an error. I fixed it. $\endgroup$ Mar 3, 2018 at 17:49
  • $\begingroup$ well, not yet: wolframalpha.com/input/… $\endgroup$
    – Wolfgang
    Mar 4, 2018 at 19:29
  • $\begingroup$ @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. $\endgroup$ Mar 4, 2018 at 19:32
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Although not as simple as nospoon's answer, here's another approach for the first one.

For $s>0$,

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


$(1)$ Euler's integral representation of the Gaussian hypergeometric function

$(2)$ Kummer's theorem

$(3)$ Duplication formula for the gamma function

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