# Integral $\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}\mathrm dx$

I need your assistance with evaluating the integral $$\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}dx$$

I tried manual integration by parts, but it seemed to only complicate the integrand more. I also tried to evaluate it with a CAS, but it was not able to handle it.

• According to WolframAlpha, the definite integral has the numerical approximation $I\approx -0.3865887494069806...$. – David H Nov 21 '13 at 2:30
• Note that the integral is equivalent to $$\frac{1}{\sqrt{2}}\int_0^{\frac{1}{\sqrt{2}}}k \log(k)K'(k) \; dk$$ where $K'(k)$ is the complementary elliptic integral of the first kind. – Shobhit Bhatnagar Nov 27 '13 at 15:47

It is easy to see that the integral is equivalent to

\begin{align*} \int_0^\infty \frac{1}{x\sqrt{2}+\sqrt{2x^2+1}}\frac{\log x}{\sqrt{1+x^2}}dx &= \sqrt{2}\int_0^\infty \frac{\sqrt{x^2+\frac{1}{2}}-x}{\sqrt{1+x^2}}\log x\; dx\tag{1} \end{align*}

This integral is a special case of the following generalised equation:

\begin{align*}\mathcal{I}(k) :&= \int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx \\ &= E'(k)-\left(\frac{1+k^2}{2} \right)K'(k)+\left(k^2 K'(k)-E'(k) \right)\frac{\log k}{2}+\log 2-1 \tag{2}\end{align*}

where $K'(k)$ and $E'(k)$ are complementary elliptic integrals of the first and second kind respectively.

Putting $k=\frac{1}{\sqrt{2}}$ in equation $(2)$,

\begin{align*} \mathcal{I}\left(\frac{1}{\sqrt{2}}\right)&=E'\left(\frac{1}{\sqrt{2} }\right)-\frac{3}{4}K'\left(\frac{1}{\sqrt{2}} \right)-\left\{\frac{1}{2} K'\left(\frac{1}{\sqrt{2}} \right)-E'\left(\frac{1}{\sqrt{2}} \right)\right\}\frac{\log 2}{4}+\log 2-1 \end{align*} Using the special values,

\begin{align*} E'\left(\frac{1}{\sqrt2} \right) &= \frac{\Gamma\left(\frac{3}{4} \right)^2}{2\sqrt\pi}+\frac{\sqrt{\pi^3}}{4\Gamma\left(\frac{3}{4} \right)^2}\\ K'\left(\frac{1}{\sqrt2} \right) &= \frac{\sqrt{\pi^3}}{2\Gamma\left(\frac{3}{4} \right)^2} \end{align*}

we get

$$\mathcal{I}\left(\frac{1}{\sqrt{2}}\right)=\frac{1+\log\sqrt[4]2}{2\sqrt{\,\pi}}\Gamma\left(\frac34\right)^2-\frac{\sqrt{\,\pi^3}}8\Gamma\left(\frac34\right)^{-2}+(\log 2-1)\, \tag{3}$$

Putting this in equation $(1)$, we get the answer that Cleo posted.

How to prove Equation $(2)$

We begin with Proposition 7.1 of this paper.

$$\int_0^\infty \frac{\log x}{\sqrt{(1+x^2)(m^2+x^2)}}dx = \frac{1}{2}K'(m)\log m$$

Multiplying both sides by $m$ and integrating from $0$ to $k$:

\begin{align*} \int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx &= \frac{1}{2}\int_0^k m K'(m)\log(m)\; dm \end{align*}

The result follows since

\begin{align*} \int m K'(m)\log(m)\; dm &= 2E'(m)-\left(1+m^2 \right)K'(m)+\left(m^2 K'(m)-E'(m) \right)\log m\\ &\quad +\text{constant} \tag{4} \end{align*}

You can verify equation $(4)$ easily by differentiating both sides with respect to $m$ and using the identities

\begin{align*} \frac{dE'(k)}{dk}&= \frac{k}{k^{'2}}(K'(k)-E'(k))\\ \frac{dK'(k)}{dk}&= \frac{k^2 K'(k)-E^{'}(k)}{kk^{'2}} \end{align*}

• I have corrected all typos/mistakes so now the answer is correct. – Shobhit Bhatnagar Nov 28 '13 at 6:18
• This is truly impressive – clathratus Jan 18 at 4:48

$$\frac{1+\ln\sqrt[4]2}{\sqrt{2\,\pi}}\Gamma\left(\frac34\right)^2-\frac{\sqrt{2\,\pi^3}}8\Gamma\left(\frac34\right)^{-2}+(\ln2-1)\,\sqrt2$$

• Would you please give some idea of how you are getting these answers? – robjohn Nov 17 '13 at 17:45
• Your creed is not in keeping with the purpose of this site, which is to help people understand. Your answers amount to numerical results with infinite precision, but offer no explanation or insight. Some people are upvoting your answers, but there are also a decent number of downvotes and flags citing that your answers are not helpful. – robjohn Nov 21 '13 at 3:13
• If you are truly as talented as your answers portray, do not waste it one being arrogant and unhelpful. Nobody not even you can benefit from this. – Ali Caglayan Nov 28 '13 at 0:00

This is not a full answer but how far I got, maybe someone can complete

$\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}=\frac{1}{\sqrt{2\,x^2}+\sqrt{2\,x^2+1}}=\frac{\sqrt{2\,x^2+1}-\sqrt{2\,x^2}}{\sqrt{2\,x^2+1}^2-\sqrt{2\,x^2}^2}=\sqrt{2\,x^2+1}-\sqrt{2\,x^2}$

So you are looking at

$$\int_0^\infty(\sqrt{\frac{2\,x^2+1}{x^2+1}} - \sqrt{\frac{2\,x^2}{x^2+1}}) \log x\,dx$$

I was then looking at $t = \frac{x^2}{x^2+1}$ but then I don't know

• this is road to perdition - both integrals are divergent – Norbert Nov 27 '13 at 12:46
• I wrote that thinking it would be clearer but you are right Norbert, thanks for spotting that. Still think it is better this way though – Thomas Nov 27 '13 at 13:34