3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
8
$\begingroup$

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

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

$\endgroup$
  • 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

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.