34
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ @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
  • 2
    $\begingroup$ 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
14
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
13
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ it is getting more and more curious ...fascinating $\endgroup$ – tired Jun 16 '15 at 19:03
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.