Who's up to the challenge? 23

Calculus Level 5

$\large \int_0^\infty e^{-x} (\ln x)^2 \, dx = A \gamma ^B + \zeta(C)$

The equation above holds true for integer constants $A,B$ and $C$ . Find $A+B+C$ .

Notations :

$\gamma$ denote the Euler-Mascheroni constant .
$\zeta(\cdot)$ denotes the Riemann zeta function .

This is a part of Who's up to the challenge?

The answer is 5.

1 solution

Aditya Narayan Sharma
May 26, 2016

Consider the integral $\displaystyle F(a)=\int_{0}^{\infty} e^{-x}x^{a-1}dx$ , where our target integral is $\displaystyle F''(1)=\int_{0}^{\infty}e^{-x}(lnx)^2dx$

We have $\displaystyle F(a)=\Gamma(a)$ , $\displaystyle F''(a)=\Gamma(a)[\psi_{0}^{2}(a)+\psi_{1}(a)]$

$\displaystyle F''(1)=\gamma^2+\frac{\pi^2}{6}=\gamma^2+\zeta(2)$ , Thus the answer is $\boxed{2+2+1=5}$

Who's up to the challenge? 23

This is a part of Who's up to the challenge?

The answer is 5.

1 solution

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