I = ∫ 0 π / 2 x lo g 2 ( sin ( x ) ) d x
Here's a simple integral for you that has a nasty result.
If I can be represented as Li 4 ( B A ) − E C π D + H lo g F ( G ) + Q π M lo g N ( P )
Find A + B + C + D + E + F + G + H + M + N + P + Q
Details and Assumptions
1) A , B , C , E , F , G , H , M , N , P , Q are positive integers not necessarily distinct, C , E are co-prime to each other, A , B are co-prime to each other , G , P are not perfect power of any integer (that it is not a perfect square, cube etc.)
2) Li s ( z ) = k = 1 ∑ ∞ k s z k . It is commonly known as Polylogarithm.
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( On printer ) +1
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If you are really printing this out, print it again, now I have corrected the typos!
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Yes, I'll be printing this, but not now. I'll try my best to understand what you're saying before I print it.
Thanks for notifying me!
I have used multiple zeta ... Rest is same evaluating that log sine integral ...
Moreover, can you explain how L s 4 ( 1 ) ( π ) evaluated in terms of λ 4 ( ) This value is directly given in that... Can you derive please ?
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The paper I quoted has indicated the necessary information to obtain the identity − L s 4 ( 1 ) ( π ) = 2 λ 4 ( 2 1 ) − 8 1 9 ζ ( 4 ) ; there is some detailed differentiation and simplification required to achieve this identity, but I think that reading the original is better than my retyping it!
Hint: Using the integral representation of H n , we have
n = 1 ∑ ∞ H n t n = − 1 − t ln ( 1 − t )
Dividing both sides by t and integrating from 0 to x , we have,
\displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n} = \operatorname{Li}_{2}(x) + \dfrac{1}{2} \ln^2(1-x) \tag{*}
Now, in the integrand, write the sin x term in ln 2 ( sin x ) as 2 i e i x − e − i x and take e i x common. Then, using the above equation and switching summation and integral, and some trivial simplifications, we have our desired equality. In between, one non trivial sum arises, i.e,
n = 1 ∑ ∞ n 2 H n = 2 ζ ( 3 )
which is left as an exercise for the reader.
This was exactly my intended solution
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I have a very long solution
I don't quite understand what you're saying. Can you explain the "trivial simplications + switching summation and integral" part?
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ln 2 ( sin x ) = [ i x + ln ( 1 − e − 2 i x ) − ln 2 i ] 2 . Expand it and use equation ( ∗ ) in the term having ln 2 ( 1 − e − 2 i x ) . Here, you have to switch sum and integral. Rest of the integrals are easily calculated using taylor series of ln ( 1 − e − 2 i x ) .
The answer is = L i 4 ( 2 1 ) − 2 8 8 0 1 9 π 4 + 2 4 lo g 4 ( 2 ) + 1 2 π 2 lo g 2 ( 2 )
So we have A = 1 , B = 2 , C = 1 9 , D = 4 , E = 2 8 8 0 , F = 4 , G = 2 , H = 2 4 , M = 2 , N = 2 , P = 2 , Q = 1 2
Where's your solution? It's been 3 MONTHS!!
Where is the solution? : )
Where is the solution? : )
it's incorrect.... how you evaluate alternating double sum as given in wikipidea
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The details of this can be found in the following: paper .
We are interested in calculating what is referred to as the log-sine integral (evaluated at π ): L s 4 ( 1 ) ( π ) = − ∫ 0 π θ lo g 2 ( 2 sin 2 1 θ ) d θ = − 4 ∫ 0 2 1 π θ ( lo g 2 + lo g ( sin θ ) ) 2 d θ , so that I = ∫ 0 2 1 π θ lo g ( sin θ ) d θ = = = = − 8 1 π 2 lo g 2 2 − 2 lo g 2 ∫ 0 2 1 π θ lo g ( sin θ ) d θ − 4 1 L s 4 ( 1 ) ( π ) 8 1 π 2 lo g 2 2 − 2 lo g 2 { 1 6 7 ζ ( 3 ) − 8 1 π 2 lo g 2 } − 4 1 L s 4 ( 1 ) ( π ) 8 1 π 2 lo g 2 2 − 8 7 lo g 2 ζ ( 3 ) − 4 1 L s 4 ( 1 ) ( π ) 8 1 π 2 lo g 2 2 − 8 7 lo g 2 ζ ( 3 ) − 3 2 1 9 ζ ( 4 ) + 2 1 λ 4 ( 2 1 ) where we are using the integral (due to Euler) ∫ 0 2 1 π θ lo g ( sin θ ) d θ = 1 6 7 ζ ( 3 ) − 8 1 π 2 lo g 2 and we have introduced a special case of the so-called Kummer polylogarithm λ 4 ( 2 1 ) = 2 k = 0 ∑ 2 k ! 1 L i n − k ( 2 1 ) lo g k 2 + 4 1 lo g 4 2 . Thus I = = = = 8 1 π 2 lo g 2 2 − 8 7 lo g 2 ζ ( 3 ) − 3 2 1 9 ζ ( 4 ) + L i 4 ( 2 1 ) + lo g 2 L i 3 ( 2 1 ) + 2 1 lo g 2 2 L i 2 ( 2 1 ) + 8 1 lo g 4 2 8 1 π 2 lo g 2 2 − 8 7 lo g 2 ζ ( 3 ) − 3 2 1 9 ζ ( 4 ) + L i 4 ( 2 1 ) + lo g 2 { − 1 2 1 π 2 lo g 2 + 6 1 lo g 3 2 + 8 7 ζ ( 3 ) } + 2 1 lo g 2 2 { 1 2 1 π 2 − 2 1 lo g 2 2 } + 8 1 lo g 4 2 L i 4 ( 2 1 ) − 3 2 1 9 ζ ( 4 ) + 2 4 1 lo g 4 2 + 1 2 1 π 2 lo g 2 2 L i 4 ( 2 1 ) − 2 8 8 0 1 9 π 4 + 2 4 1 lo g 4 2 + 1 2 1 π 2 lo g 2 2 so that the answer is 1 + 2 + 1 9 + 4 + 2 8 8 0 + 4 + 2 + 2 4 + 2 + 2 + 2 + 1 2 = 2 9 5 4 .