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Calculus Level 5

0 1 1 + x 13 d x = a b π csc ( π c ) \large \displaystyle\int _{ 0 }^{ \infty }{ \dfrac { 1 }{ 1+{ x }^{ 13 } } \, dx } =\dfrac { a }{ b } \pi \csc { \left( \dfrac { \pi }{ c } \right) }

If the equation above holds true for positive integers a , b a,b and c c , with a , b a,b coprime, find a + b + c a+b+c .


The answer is 27.

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2 solutions

Parv Mor
Apr 25, 2016

One can use contour Integration to show that x m / ( 1 + x n ) d x = ( π / n ) / s i n ( m + 1 ) π / n . \int x^m/(1+x^n)dx=(π/n)/sin(m+1)π/n. with the given set of limits(zero-infinite).For us the integral is simply π / 13 c s c ( π / 13 ) π/13csc(π/13)

Spandan Senapati - 4 years, 1 month ago
First Last
Oct 15, 2016

A beautiful solution I have seen multiple times on Math Stack:

0 1 x n + 1 d x = 0 0 e t ( 1 + x n ) d x d t \displaystyle\int_{0}^{\infty}\frac{1}{x^n+1}dx = \int_{0}^{\infty}\int_{0}^{\infty}e^{-t(1+x^n)}dxdt

Sub u = x n t d u = n x n 1 t d x \displaystyle u = x^nt\quad du = nx^{n-1}t\,dx

1 n 0 e t t 1 n d t 0 u 1 n 1 e u d u = \displaystyle\frac{1}{n}\int_{0}^{\infty}e^{-t}t^{\frac{-1}{n}}dt\int_{0}^{\infty}u^{\frac{1}{n}-1}e^{-u}du =

Γ ( 1 1 n ) Γ ( 1 n ) n = π n sin π n by Euler’s Reflection Formula \displaystyle\frac{\Gamma(1-\frac{1}{n})\Gamma(\frac{1}{n})}{n} = \frac{\pi}{n\sin{\frac{\pi}{n}}}\quad \text{by Euler's Reflection Formula}

Plug in 13 for π 13 sin π 13 \displaystyle\huge{\color{#D61F06}\frac{\pi}{\color{#3D99F6}13\color{#20A900}\sin{\color{#EC7300}\frac{\pi}{13}}}}

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