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Calculus Level 5

$\int_{0}^{\infty}\sin^{-1}\left(\dfrac{e^{-x}}{\sqrt{2}}\right)\, dx=\dfrac{\pi^P}Q \ln R + \dfrac GR$

The equation above holds true for positive integers $P,Q,R$ with $R$ being a prime. Find $P+Q+R$

Notation: $\displaystyle G = \sum_{n=0}^\infty \dfrac{ (-1)^n}{(2n+1)^2} \approx 0.916$ denotes the Catalan's constant .

Try this first .

The answer is 11.

1 solution

Chew-Seong Cheong
Nov 26, 2016

$\begin{aligned} I & = \int_0^\infty \left(\frac {e^{-x}}{\sqrt 2} \right) \ dx & \small \color{#3D99F6} \text{Let } \sin \theta = \left(\frac {e^{-x}}{\sqrt 2} \right); \ \cos \theta \ d \theta = - \left(\frac {e^{-x}}{\sqrt 2} \right) \ dx \\ & = \int_0^\frac \pi 4 \theta \cot \theta \ d \theta & \small \color{#3D99F6} \text{By integration by parts} \\ & = \theta \ln (\sin \theta) \bigg|_0^\frac \pi 4 - \int_0^\frac \pi 4 \ln (\sin \theta) \ d \theta \\ & = \frac \pi 4 \ln \left( \frac 1{\sqrt 2} \right) - {\color{#3D99F6} \int_0^\frac \pi 4 \ln (2\sin \theta) \ d \theta} + \int_0^\frac \pi 4 \ln 2 \ d \theta \\ & = - \frac {\pi \ln 2}8 + {\color{#3D99F6} \frac G2} + \frac {\pi \ln 2}4 & \small \color{#3D99F6} G\text{ is the Catalan's constant.} \\ & = \frac {\pi \ln 2}8 + \frac G2 \end{aligned}$

$\implies P+Q+R =1 + 8 + 2 = \boxed{11}$ .

This needs more care!

Try this first .

The answer is 11.

1 solution

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