I just found on Mathematica that the solution to the integral $$\int_{0}^{\infty} \log(x) \cdot K_{0}(2\sqrt{x})\,dx$$ is actually minus Euler–Mascheroni constant, where $K_{0}(\cdot)$ is the modified Bessel function of second kind.

This is very interesting given that $\int_{0}^{\infty} \log(x) \cdot \exp(-x)\,dx$ also equals to minus Euler constant.

However, I did not find any sources which has this equality and I am getting a little bit suspicious about this equality. I tried to use the series expansion of modified Bessel function of 2nd kind for the integral, but I still cannot get the answer, so do you guys have any brilliant ideas to prove the equality.

Thanks a lot.

  • $\begingroup$ In fact both integrals give $\color{red}{-}\gamma$ $\endgroup$ – Claude Leibovici Jan 29 '18 at 5:12
  • $\begingroup$ @ClaudeLeibovici, yes, it should be minus $-\gamma$, it is now corrected in the text. thanks for pointing it out. :) $\endgroup$ – Vic Jan 29 '18 at 8:44

I give you an outline that should work, and leave it to you to fill in details. The Mellin transform of $K_0$ can be calculated, $$ \int_0^{+\infty}t^{s-1}K_0(t)\,dt=2^{s-2}\Gamma(s/2)^2. $$ Your integral can after the substitution $t=2\sqrt{x}$ be written as $$ \int_0^{+\infty}t\ln (t/2) K_0(t)\,dt. $$ If we differentiate the identity from the Mellin transform with respect to $s$, and set $s=2$, we find that $$ \int_0^{+\infty} t\ln t K_0(t)\,dt=-\gamma+\ln 2. $$ Using $\int_0^{+\infty}t K_0(t)\,dt=1$, you will get

$$\int_0^{+\infty} \ln (x) K_0(2\sqrt{x})\,dx=-\gamma.$$

  • 1
    $\begingroup$ that´s really an elegant and brilliant way to prove the equality, thanks a lot :). $\endgroup$ – Vic Jan 29 '18 at 8:41
  • $\begingroup$ hi, mickep, while I am playing with a mathematical model, an integral involving K function occurs, could you have a look and see if you have any thoughts (bit.ly/2qael0y)? Thanks a lot in advance. $\endgroup$ – Vic Apr 7 '18 at 1:10

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