Who's up to the challenge? 2

Calculus Level 5

$\large \int_0^\infty \dfrac{x^9 \ln x}{e^x} \, dx$

If the above integral can be expressed as $144(B -C\gamma),$ where $B$ and $C$ are integers, find $B+C+144$ .

Notation: $\gamma$ denotes the Euler-Mascheroni constant $\displaystyle \gamma = \lim_{n\to\infty} \left( - \ln n + \sum_{k=1}^n \dfrac1k \right) \approx 0.5772 .$

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The answer is 9793.

1 solution

Aareyan Manzoor
Feb 18, 2016

$\Gamma(n)=\int_0^\infty x^{n-1}e^{-x}\text{dx}\\ \Gamma'(n)=\int_0^\infty \ln(x)x^{n-1}e^{-x}\text{dx}\\ \text{given integral}= \Gamma'(10)=\Gamma(10)\psi(10)\\ =9!\left(\psi(1)+H_9\right)=9!(H_9-\gamma)=144(7129-2520\gamma)\\ \therefore 144+7129+2520=\boxed{9793}$

exactly as intended!

Hamza A - 5 years, 3 months ago

Sweet indeed

Parth Lohomi - 5 years ago

Who's up to the challenge? 2

The answer is 9793.

1 solution

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