Who's up to the challenge?22

Calculus Level 5

∫ 0 ∞ x sinh ⁡ x e x 2 d x = e A ⋅ π B C \displaystyle\int _{ 0 }^{ \infty }{ \frac { x\sinh { x } }{ { e }^{ { x }^{ 2 } } } dx } =\frac { \sqrt [ A ]{ e } \cdot \sqrt [ B ]{ \pi } }{ C } \\

where A , B , C A,B,C are positive integers

Find A + B + C A+B+C

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


The answer is 10.

Hamza A by Hamza A , 19, USA

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1 solution

First Last
Jun 9, 2017

I = 1 2 ∫ R x sinh ⁡ x e x 2 d x = 1 4 ∫ R x − x 2 + x − x e − x 2 − x d x sinh ⁡ x = e x − e − x 2 and R is the real number line \displaystyle I = \frac1{2}\int_R\frac{x\sinh{x}}{e^{x^2}}dx=\frac1{4}\int_R x^{-x^2+x}-xe^{-x^2-x}dx\quad\sinh{x}=\frac{e^x-e^-x}{2}\text{ and R is the real number line}

∫ R x e − ( x + b ) 2 d x = ∫ R ( t − b ) e − t 2 d t = 0 − b π \displaystyle\int_R xe^{-(x+b)^2}dx=\int_R (t-b)e^{-t^2}dt= 0 - b\sqrt{\pi}

Using this I becomes I = 1 4 ∫ R x e − ( x − 1 2 ) 2 + 1 4 − x e − ( x + 1 2 ) 2 + 1 4 d x = e 1 4 π 4 \displaystyle I = \frac1{4}\int_R xe^{-(x-\frac1{2})^2+\frac1{4}}-xe^{-(x+\frac1{2})^2+\frac1{4}}dx = \boxed{\frac{e^\frac1{4}\sqrt{\pi}}{4}}

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So what does that mean....

. . - 3 months, 3 weeks ago

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