How will you solve this?

Calculus Level 5

$\large \displaystyle \int_0 ^ \infty e^{-x^2} \ln(x)\ \mathrm{d}x = -\dfrac{A}{B}\left( \gamma^C +D\ln(D) \right)\pi^E$

If the above equation is true for positive integers $A,B,C,D$ , where $A,B$ are coprime to each other and $D$ isn't any $m^{th}$ power of a positive integer with $m \in \mathbb Z,\ m \geq 2$ .

Submit the value of $A+B+C+D+2E$ as your answer.

Note: $\gamma$ is the Euler-Mascheroni Constant defined as:

$\large \gamma =\lim_{n \rightarrow \infty} \left(\displaystyle \sum_{k=1} ^{n} \dfrac{1}{k} -\ln(n) \right)$

The answer is 9.

1 solution

Kunal Gupta
Sep 27, 2015

Consider the integral, $I(a)= \displaystyle \int_{0}^{\infty}x^{a}e^{-x^2}dx$ Setting, $x^2 \rightarrow x$ $I(a)= \dfrac{1}{2}\displaystyle \int_{0}^{\infty}x^{\frac{a-1}{2}}e^{-x}dx = \dfrac{1}{2}\Gamma\left(\dfrac{a+1}{2}\right)$ Clearly, our required integral is the following when $a=0$ $\dfrac{\partial I(a)}{\partial a} =\displaystyle \int_{0}^{\infty}x^{a}\ln{(x)}e^{-x^2}dx$
$I'(a)= \dfrac{1}{4}\Gamma'\left(\dfrac{a+1}{2}\right) = \dfrac{1}{4}\Gamma\left(\dfrac{a+1}{2}\right)\psi\left(\dfrac{a+1}{2}\right)$ where $\psi(n)$ is the Digamma Function Plugging $a=0$ $I'(0) = \dfrac{1}{4}\sqrt{\pi}\psi(\frac{1}{2})$ Using the fact that $\psi(1/2) = -2\ln(2) - \gamma$ $I'(0) = -\dfrac{1}{4}\left( \gamma +2\ln(2) \right)\sqrt{\pi}$

I too did exactly in the same way.

Surya Prakash - 5 years, 8 months ago

How will you solve this?

The answer is 9.

1 solution

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