0 ∫ ∞ e x − 1 lo g 2 ( 1 − e − x ) x 5 d x
Let 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 )
Evaluate A + B + C + D + E + F + G + H + J + K
∙ A , B , . . . . , K all in positive integers.
■ This is a part of Hot Integrals
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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 ( e x − 1 ) n x 5 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.
Here is my solution.
I won't be writing it again. But yeah, I will give the answer form -
2 0 π 2 ζ ( 3 ) 2 − 7 2 0 ζ ( 3 ) ζ ( 5 ) + 6 1 π 2 ζ ( 6 )
@Aman Raimann Really thanks for the problem! Enjoyed solving it! I wasted a lot of time in computing the Euler sums though :/
solution , i dont like that much ! :/ but yeah,.!! Congrats for solving dude :) :) :) :)
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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.
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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
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Put e − x = t . You will get , 1 ∫ 0 1 − t lo g 2 ( 1 − t ) lo g 5 t d t
which can further be written as − 0 ∫ 1 t lo g 2 ( t ) lo g 5 ( t ) d t
this famous integral is polylog function, and is equivalent to
S 3 , 5 ( 1 ) Γ ( 3 ) Γ ( 6 )
which is equal to
2 0 π 2 ζ 2 ( 3 ) − 7 2 0 ζ ( 3 ) ζ ( 5 ) + 6 1 π 2 ζ ( 6 )