40
$\begingroup$

I discovered the following conjecture by evaluating the integral numerically and then using some inverse symbolic calculation methods to find a possible closed form: $$\int_0^\infty\frac{\ln x}{\sqrt{x\vphantom{1}}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel{\color{#808080}?}=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\,\Gamma^2\left(\tfrac34\right)}.\tag1$$ The equality holds numerically with a precision of at least $1000$ decimal digits. But so far I was not able to find a proof of it.

Because the integral can be represented as a derivative of a hypergeometic function with respect to its parameter, the conjecture can be rewritten as $$\frac{d}{da}{_2F_1}\left(a,\ \tfrac12;\ 1;\ \tfrac12\right)\Bigg|_{a=\frac12}\stackrel{\color{#808080}?}=\frac{\sqrt\pi\,\ln2}{2\,\Gamma^2\left(\tfrac34\right)}\tag2$$ or, using a series expansion of the hypergeometric function, as $${\large\sum}_{n=0}^\infty\frac{H_{n-\frac12}\ \Gamma^2\left(n+\tfrac12\right)}{2^n\ \Gamma^2\left(n+1\right)}\stackrel{\color{#808080}?}=-\frac{3\,\pi^{3/2}\,\ln2}{2\,\Gamma^2\left(\tfrac34\right)}\tag3,$$ where $H_q$ is the generalized harmonic number, $H_q=\gamma+\psi_0\left(q+1\right).$

Could you suggest any ideas how to prove this?

$\endgroup$
2
  • 3
    $\begingroup$ You seem to have forgotten the minus sign, since the quantity is negative. Also, letting $x=\sinh^2t$, the integral can be rewritten as $\displaystyle\int_0^\infty\frac{\ln(\sinh t)}{\sqrt{\cosh(2t)}}dt$. Perhaps something similar exists in Gradshteyn-Ryzhik ? $\endgroup$
    – Lucian
    Commented Jul 25, 2014 at 9:10
  • $\begingroup$ @Lucian they doesn’t looks same numerically $\endgroup$ Commented Mar 11 at 18:06

1 Answer 1

50
$\begingroup$

$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$

After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$:

$$I=\int_{0}^{\infty}\frac{2\ln{(x)}}{\sqrt{2x}\,\sqrt{2x+2}\,\sqrt{2x+1}}\mathrm{d}x=\int_{0}^{\infty}\frac{\ln{\left(\frac{t}{2}\right)}}{\sqrt{t}\,\sqrt{t+2}\,\sqrt{t+1}}\mathrm{d}t.$$

Next, substituting $t=\frac{1}{u}$ yields:

$$\begin{align} I &=-\int_{0}^{\infty}\frac{\ln{(2u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-\int_{0}^{\infty}\frac{\ln{(u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-I\\ \implies I&=-\frac{\ln{(2)}}{2}\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}. \end{align}$$

Making the sequence of substitutions $x=\frac{u-1}{2}$, then $u=\frac{1}{t}$, and finally $t=\sqrt{w}$, puts this integral into the form of a beta function:

$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}w}{w^{3/4}\,\sqrt{1-w}}\\ &=\frac12\operatorname{B}{\left(\frac14,\frac12\right)}\\ &=\frac12\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac14\right)}}{\Gamma{\left(\frac34\right)}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}} \end{align}$$

Hence,

$$I=-\frac{\ln{(2)}}{2}\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}=-\frac{\pi^{3/2}\,\ln{(2)}}{2^{3/2}\,\Gamma^2{\left(\frac34\right)}}.~~~\blacksquare$$


Possible Alternative: You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=\sqrt{w}$.

$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\ &=2\,K{(-1)}\\ &=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}. \end{align}$$

$\endgroup$
2
  • 10
    $\begingroup$ Very nice answer. +1 After about 30 minutes of getting nowhere, I gave up. $\endgroup$ Commented Jul 25, 2014 at 22:05
  • 3
    $\begingroup$ @RandomVariable Thanks. That compliment means a lot coming from you. $\endgroup$
    – David H
    Commented Jul 25, 2014 at 22:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .