Hot Integral - 4

Calculus Level 5

0 log 2 ( 1 e x ) x 5 e x 1 d x \large\displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx

Let I I denotes the above integral which can be expressed the form of

I = A π B ζ C ( D ) E ζ ( F ) ζ ( G ) + H π J ζ ( K ) \displaystyle I = A\pi^B\zeta^C(D) - E\zeta(F)\zeta(G) + H\pi^J\zeta(K)

Evaluate A + B + C + D + E + F + G + H + J + K A+B+C+D+E+F+G+H+J+K

A , B , . . . . , K \bullet A,B,....,K all in positive integers.

\blacksquare This is a part of Hot Integrals


The answer is 824.

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2 solutions

Aman Rajput
Jul 24, 2015

Put e x = t e^{-x}=t . You will get , 1 0 log 2 ( 1 t ) log 5 t 1 t d t \displaystyle \int\limits_1^0 \frac{\log^2(1-t)\log^5t}{1-t} dt

which can further be written as 0 1 log 2 ( t ) log 5 ( t ) t d t \displaystyle -\int\limits_0^1 \frac{\log^2(t)\log^5(t)}{t} dt

this famous integral is polylog function, and is equivalent to

S 3 , 5 ( 1 ) Γ ( 3 ) Γ ( 6 ) \displaystyle S_{3,5}(1)\Gamma(3)\Gamma(6)

which is equal to

20 π 2 ζ 2 ( 3 ) 720 ζ ( 3 ) ζ ( 5 ) + 61 π 2 ζ ( 6 ) 20\pi^2\zeta^2(3) - 720\zeta(3)\zeta(5) + 61\pi^2\zeta(6)

This could be done first by multiplying diving the log internally by e^x. Then seperate the integrals. Now in the first integral u can consider it as the second derivative of x 5 ( e x 1 ) n \frac{x^5}{(e^x-1)^n} wrt n at n=1. Then we can proceed further easily. But there will be a bit problem in getting the solution to the form asked here.

Aditya Kumar - 5 years, 4 months ago
Kartik Sharma
Jul 22, 2015

Here is my solution.

I won't be writing it again. But yeah, I will give the answer form -

20 π 2 ζ ( 3 ) 2 720 ζ ( 3 ) ζ ( 5 ) + 61 π 2 ζ ( 6 ) \displaystyle 20{\pi}^{2}{\zeta(3)}^{2} - 720\zeta(3)\zeta(5) + 61{\pi}^{2}\zeta(6)

@Aman Raimann Really thanks for the problem! Enjoyed solving it! I wasted a lot of time in computing the Euler sums though :/

Kartik Sharma - 5 years, 10 months ago

solution , i dont like that much ! :/ but yeah,.!! Congrats for solving dude :) :) :) :)

Aman Rajput - 5 years, 10 months ago

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You have a better one? Post yours! I truly loved solving it but yeah failed in my job to impress :/

Still, I think it gives a general approach at least.

Kartik Sharma - 5 years, 10 months ago

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Ya .. i have something interesting for that ..,! i will post it tomorrow ! Now its 2:00 AM cant open laptop! :D

Aman Rajput - 5 years, 10 months ago

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