How to prove the following conjectured identity? $$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$ It holds numerically with precision of at least $1000$ decimal digits.

Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?

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    $\begingroup$ I've reduced it to $$2^{1/4} \int_0^{\infty} \frac{dv}{(v^2+8)^{1/4} (v^2+2)^{1/2}} $$ $\endgroup$ – Ron Gordon Jun 15 '15 at 19:01
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    $\begingroup$ It is interesting to point out that Mathematica evaluates the equivalent $$\int_{0}^{1}\frac{dx}{x^{3/4}(1-x)^{1/2}(x+1/3)^{1/4}}$$ almost instantly. $\endgroup$ – Jack D'Aurizio Jun 15 '15 at 19:17
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    $\begingroup$ may this be related to the fact that $K(\frac{\sqrt{2}}{2})=\frac{1}{4\sqrt{\pi}}\Gamma^2(\frac{1}{4})$ where $K$ is an elliptic integral of type one. $\endgroup$ – tired Jun 15 '15 at 19:24
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    $\begingroup$ @tired Here a similarly looking identity $$ \int_0^{\infty} \frac{dx}{ \sqrt[3]{55+\cosh x}} = \frac{\sqrt[3]2\,\sqrt3}{7\pi} \Gamma^3\!\!\left(\tfrac13\right)$$ was proved. $\endgroup$ – Start wearing purple Jun 15 '15 at 21:49
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    $\begingroup$ Looks like $7$ can be replaced by $161$, and also $0$ and a few rational values like $5/4$ and $65/16$ and irrationalities like $\sqrt 5$ and $5 \sqrt{13}$. But I need to get some sleep before I write more; I know how to do this but Jack d'Aurizio underestimates the time required by about three orders of magnitude... $\endgroup$ – Noam D. Elkies Jun 20 '15 at 5:21

I will follow @user15302's idea. In this answer, I showed that

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

where $b = \frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = \frac{1}{4}$ and $a = 7$. Then we have $b = \frac{3}{4}$ and

$$ I = 2^{-3/4} \int_{0}^{1} \frac{1}{v^{3/4}\sqrt{(1-v)(1-\frac{3}{4}v)}} \, dv. $$

The reason why the case $b = \frac{3}{4}$ is special is that, if we plug $v = \operatorname{sech}^2 t$ then we can utilize the triple angle formula to get the following surprisingly neat integral

$$ I = 2^{5/4} \int_{0}^{\infty} \frac{\cosh t}{\sqrt{\cosh 3t}} \, dt. $$

Now using the substitution $x = e^{-6t}$, we easily find that

$$ I = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{1} \frac{x^{-11/12} + u^{-7/12}}{\sqrt{1+x}} \, dx = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{\infty} \frac{dx}{x^{11/12}\sqrt{1+x}}. $$

The last integral can be easily calculated by the following formula

$$ \int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}} \, dx = \beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$

Therefore we obtain the following closed form

$$ I = \frac{\Gamma(\frac{1}{12})\Gamma(\frac{5}{12})}{3 \sqrt[4]{2}\sqrt{\pi}}. $$

In order to verify that this is exactly the same as Vladimir's result, We utilize the Legendre multiplication formula and the reflection formula to find that

$$ \Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12}) = \frac{\Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})\Gamma(\tfrac{9}{12})}{\Gamma(\tfrac{3}{4})} = \frac{2 \pi \cdot 3^{1/4} \Gamma(\frac{1}{4})}{\Gamma(\tfrac{3}{4})} = 2^{1/2} 3^{1/4} \Gamma(\tfrac{1}{4})^2. $$

This completes the proof.

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    $\begingroup$ Wonderful. Really well done! $\endgroup$ – nospoon Jul 31 '15 at 6:58
  • $\begingroup$ @Sanchul: Perhaps this related question may be of interest? $\endgroup$ – Tito Piezas III Dec 5 '16 at 4:29

By replacing $x$ with $4u$, then $\cosh u$ with $\frac{1}{t}$, we have:

$$ I = \frac{1}{2^{3/4}}\int_{0}^{1}\frac{dt}{(1-t^2+t^4)^{1/4}(1-t^2)^{1/2}}=\frac{1}{2^{3/4}}\int_{0}^{1/2}\frac{dt}{(1-t(1-t))^{1/4}(t(1-t))^{1/2}} $$ Next, by replacing $t(1-t)$ with $v/4$, $$ I=\frac{1}{2^{11/4}}\int_{0}^{1}\frac{dv}{(1-v/4)^{1/4}(v(1-v))^{1/2}}$$ then, by setting $v=4-3z$, $$ I = \frac{3^{1/4}}{2^{9/4}}\int_{1}^{4/3}\frac{dz}{z^{1/4}((4-3z)(1+z))^{1/2}}=\frac{3^{1/4}}{2^{5/4}}\int_{1}^{2/\sqrt{3}}\sqrt{\frac{z}{(4-3z^2)(1+z^2)}}\,dz$$ that, at least, looks manageable. We also have: $$ I = \frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{(3u^4+u^2)^{1/4}(1-u^2)^{1/2}}\tag{1}$$ that Mathematica gladly evaluates to: $$ I = \frac{2^{1/4}\,\Gamma\left(\frac{1}{4}\right)^2}{3^{3/4}\sqrt{\pi}}. $$ Now we just need to understand how.

I think this problem can be solved by invoking the theory of $j$-invariants for (hyper?)-elliptic curves, but I am not so confident in the topic to find the right change of variables that brings our integral into a complete elliptic integral. I think that Noam Elkies would solve this problem in a few seconds, so I am asking his help.

Update. Found. Our claim was proven by Zucker and Joyce in Special values of the hypergeometric series II, it is the result $(7\!\cdot\! 6)$. It is derived through standard hypergeometric manipulations, by starting with the elliptic modulus $k$ for which: $$\frac{K'(k)}{K(k)}=3.$$ The modular function to be considered for regarding our integral as a period is so the elliptic lambda function.

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    $\begingroup$ I suppose you know that, but performing a substitution $u=\sqrt{q}$ your last integral can be brought in contact with the hypergeometric $_2F_1$. Maybe one of their multiple transformation properties can finish it off, by relating it with for example the elliptic integral i posted above. $\endgroup$ – tired Jun 16 '15 at 8:53
  • $\begingroup$ This approach suggests that the question is equivalent to show that $_2F_1(\frac{1}{4},\frac{1}{4},\frac{3}{4},-3)=\frac{2}{3^{3/4}}$, can somebody bring it home from here? $\endgroup$ – tired Jun 16 '15 at 9:11
  • $\begingroup$ Can you please show how you got the integral representation marked $(1)$? Thanks! $\endgroup$ – Pranav Arora Jun 16 '15 at 13:43
  • $\begingroup$ @PranavArora : $(1)$ follows from replacing $x$ with $2u$, then exploiting $7+T_2(x) = 2(x^2+3)$. $\endgroup$ – Jack D'Aurizio Jun 16 '15 at 13:56
  • $\begingroup$ The 2 in the denominator of the RHS should be a 3. Edit suggested... $\endgroup$ – John Molokach Jun 16 '15 at 14:07

Take the integral in the form $$I=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{4}+u^{2}\right)^{1/4}\left(1-u^{2}\right)^{1/2}}=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{2}+1\right)^{1/4}\left(1-u^{2}\right)^{1/2}\left(u^{2}\right)^{1/4}} $$ then put $u^{2}=s $ $$=\frac{1}{2^{5/4}}\int_{0}^{1}\frac{ds}{\left(3s+1\right)^{1/4}\left(1-s\right)^{1/2}s^{3/4}} $$ and now put $s=1-t $ $$=\frac{1}{2^{7/4}}\int_{0}^{1}\frac{dt}{\left(1-3t/4\right)^{1/4}\left(1-t\right)^{3/4}t^{1/2}}. $$ Now recalling the identity $$\,_{2}F_{1}\left(a,b;c;z\right)=\frac{\Gamma\left(c\right)}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-tz\right)^{a}}dt $$ we have $$I=\frac{1}{2^{7/4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right) $$ and in this case it is possible calculate the exact value of the hypergeometric function (see the update in the Jack D'Aurizio's answer for reference) $$\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)=\frac{2\sqrt{2}}{3^{3/4}} $$ and so $$I=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6^{3/4}\sqrt{\pi}}. $$ The result is not equal to $\sqrt[4]{6}\Gamma^{2}\left(\frac{1}{4}\right)/\left(2\sqrt{\pi}\right) $ but I haven't found a mistake in my calculations.

  • $\begingroup$ this approach is up to a linear transformation the same which i sketched below @Jack d'Aurizio's answer. Do you have a proof of how to get the exact value for the hypergeometric? $\endgroup$ – tired Jun 16 '15 at 10:02
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    $\begingroup$ @tired No, but I think it is from the identities involving $\,_{2}F_{1}\left(a,b;a-b+1;z\right).$ $\endgroup$ – Marco Cantarini Jun 16 '15 at 10:24
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    $\begingroup$ @JackD'Aurizio the same is true for "my" representation $_2F_1(\frac{1}{4},\frac{1}{4},c,z)$. The value for $c=\frac{-3}{4}$ is elementary but $\frac{3}{4}$ is missing completly. functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… $\endgroup$ – tired Jun 16 '15 at 14:51
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    $\begingroup$ @tired: that is not surprising, since the two representations are the same up to Pfaff transformation. $\endgroup$ – Jack D'Aurizio Jun 16 '15 at 15:03
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    $\begingroup$ it is getting more and more curious ...fascinating $\endgroup$ – tired Jun 16 '15 at 19:03

This is a partial answer to the second question. Mathematica could evaluate $$\int_0^\infty\frac{dx}{\sqrt[4]{a+\cosh x}},$$ in term of the following Appell function: $$ F_1\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{5}{4},\sqrt{a^2-1}-a,\frac{1}{\sqrt{a^2-1}-a}\right). $$ For $a=0$ and $a=1$ there is a closed-form of this Appell function, so we get $$\int_0^\infty\frac{dx}{\sqrt[4]{\cosh x}} = \frac{4\sqrt{\pi}\,\Gamma\left(\frac{9}{8}\right)}{\Gamma\left(\frac{5}{8}\right)}$$ and $$\int_0^\infty\frac{dx}{\sqrt[4]{1+\cosh x}} = \frac{\Gamma^2\left(\frac{1}{4}\right)}{2^{3/4}\sqrt{\pi}}.$$ Numerically I've got your conjectured form for $a=7$ too: $$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\left(\tfrac14\right),$$ or in term of elliptic $K$ function: $$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{4\sqrt[4]2\sqrt[4]3}{3}K\left(\tfrac{\sqrt{2}}{2}\right).$$

A related, somehow generalized question is here.


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