Are they nested?

Calculus Level 5

$\Large \int_0^\infty \frac {x^3}{e^x - \frac {\sum_{k=1}^\infty \frac 1{k+12k^2}}{\int_0^{\pi/2} \big(2\tan \theta - 2\tan \theta \sin^{1/6} \theta \big)\ d\theta}} dx = \frac 1{A\pi^B}$

Given that $A$ and $B$ are integers, find the value of $A+B$ .

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The answer is 11.

1 solution

Aditya Kumar
Nov 21, 2015

$I=\int _{ 0 }^{ \infty }{ \left( \frac { { x }^{ 3 } }{ { e }^{ x }-\left( \frac { \sum _{ a=1 }^{ \infty }{ \left( \frac { 1 }{ a+12{ a }^{ 2 } } \right) } }{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2tan\left( \alpha \right) -2tan\left( \alpha \right) { \left( sin\left( \alpha \right) \right) }^{ \frac { 1 }{ 6 } } \right) } d\alpha } \right) } \right) } dx\\ { I }_{ 1 }=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2tan\left( \alpha \right) -2tan\left( \alpha \right) { \left( sin\left( \alpha \right) \right) }^{ \frac { 1 }{ 6 } } \right) } d\alpha \\ Let\quad { \left( sin\left( \alpha \right) \right) }^{ 2 }=y\\ \therefore \quad { I }_{ 1 }=\int _{ 0 }^{ 1 }{ \left( \frac { 1-{ y }^{ \frac { 1 }{ 12 } } }{ 1-y } \right) \\ } \quad =\quad { H }_{ \frac { 1 }{ 12 } }\\ S=\sum _{ a=1 }^{ \infty }{ \left( \frac { 1 }{ a+12{ a }^{ 2 } } \right) } \\ \quad =\quad \frac { 1 }{ 12 } \sum _{ a=1 }^{ \infty }{ \left( \frac { 1 }{ a\left( \frac { 1 }{ 12 } +a \right) } \right) } =\quad { H }_{ \frac { 1 }{ 12 } }\\ \therefore \quad I=\int _{ 0 }^{ \infty }{ \left( \frac { { x }^{ 3 } }{ { e }^{ x }-1 } \right) } dx\\ \quad \quad \quad =\int _{ 0 }^{ \infty }{ \left( \frac { { x }^{ 4-1 } }{ { e }^{ x }-1 } \right) } dx\\ \quad \quad \quad \quad \left\{ M\quad \frac { 1 }{ { e }^{ x }-1 } \right\} \left( 3 \right) ,\\ Here\quad \left\{ M\quad f \right\} \left( s \right) \quad is\quad mellin\quad transform\quad of\quad f\quad over\quad s.\\ \therefore \quad By\quad Ramanujan's\quad master\quad theorem,\\ \quad \quad I=\Gamma \left( 4 \right) \zeta \left( 4 \right) =\frac { { \pi }^{ 4 } }{ 15 } =\frac { 1 }{ 15{ \pi }^{ -4 } } \\ A=15\quad B=-4\quad \\ \therefore \quad A+B=11$

I too did in the same way. But at last I did like this.

$\begin{aligned} \int_{0}^{\infty} \dfrac{x^3}{e^x-1} &= \int_{0}^{\infty} \dfrac{x^3 e^{-x}}{1-e^{-x}} \\ &= \int_{0}^{\infty} x^{3} \sum_{r=1}^{\infty} e^{-rx} \\ &= \sum_{r=1}^{\infty} \int_{0}^{\infty}x^3 e^{-rx} dx \\ &=\sum_{r=1}^{\infty} \dfrac{1}{r^4} \int_{0}^{\infty} t^3 e^{-t} dt \\ &= \Gamma (4) \times \zeta (4) = \dfrac{\pi^4}{15} \end{aligned}$

Surya Prakash - 5 years, 6 months ago

Yeah! Thumbs up for simplicity!

Pi Han Goh - 5 years, 6 months ago

Thumbs up