Primes everywhere

Calculus Level 5

∫ 0 ∞ 1 x 13 + 13 d x = 1 a b / a ∫ 0 ∞ 1 1 + x 13 d x \displaystyle\int _{ 0 }^{ \infty }{ \dfrac { 1 }{ { x }^{ 13 }+13 } \, dx } =\dfrac { 1 }{ { a }^{ b/a } } \int _{ 0 }^{ \infty }{ \dfrac { 1 }{ 1+{ x }^{ 13 } } \, dx }

The equation above holds true for constants a a and b b with prime a a . Find a + b a+b .


The answer is 25.

Hamza A by Hamza A , 19, USA

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1 solution

First Last
Jun 14, 2017

∫ 0 ∞ 1 1 + x 13 d x = ∫ 0 ∞ ∫ 0 ∞ e − t ( 1 + x 13 ) d t d x = \displaystyle\int_0^\infty\frac1{1+x^{13}}dx=\int_0^\infty\int_0^\infty e^{-t(1+x^{13})}dt\,dx=

1 13 ∫ 0 ∞ e − u u − 12 13 d u ∫ 0 ∞ e − t t − 1 13 d t by u = t x 13 \frac1{13}\int_0^\infty e^{-u}u^\frac{-12}{13}du\int_0^\infty e^{-t}t^\frac{-1}{13}dt\quad\text{by }u=tx^{13}

Using the same idea: ∫ 0 ∞ 1 13 + x 1 3 d x = ∫ 0 ∞ ∫ 0 ∞ e − t ( 13 + x 13 ) d t d x = 1 13 ∫ 0 ∞ e − u u − 12 13 d u ∫ 0 ∞ e − 13 t t − 1 13 d t \displaystyle\int_0^\infty\frac1{13+x^13}dx=\int_0^\infty\int_0^\infty e^{-t(13+x^{13})}dt\,dx = \frac1{13}\int_0^\infty e^{-u}u^\frac{-12}{13}du\int_0^\infty e^{-13t}t^\frac{-1}{13}dt

And w = 13 t w=13t results in:

∫ 0 ∞ 1 13 + x 13 d x = 1 3 1 13 1 3 2 ∫ 0 ∞ e − u u − 12 13 d u ∫ 0 ∞ e − w w − 1 13 d w = 1 1 3 12 13 ∫ 0 ∞ 1 1 + x 13 d x \displaystyle\int_0^\infty\frac1{13+x^{13}}dx = \frac{13^\frac{1}{13}}{13^2}\int_0^\infty e^{-u}u^\frac{-12}{13}du\int_0^\infty e^{-w}w^\frac{-1}{13}dw = \frac1{13^\frac{12}{13}}\int_0^\infty\frac1{1+x^{13}}dx

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