Simple looking integrals can be very nasty

Calculus Level 5

I = 0 π / 2 x log 2 ( sin ( x ) ) d x I = \displaystyle \int _{ 0 }^{ \pi /2 }{ x \ { \log }^{ 2 }(\sin(x)) \ \mathrm{d}x }

Here's a simple integral for you that has a nasty result.

If I I can be represented as Li 4 ( A B ) C π D E + log F ( G ) H + π M log N ( P ) Q { \text{Li} }_{ 4 }\left(\dfrac { A }{ B } \right)-\dfrac { C{ \pi }^{ D } }{ E } +\dfrac { { \log }^{ F }(G) }{ H } +\dfrac { { \pi }^{ M }{ \log }^{ N }(P) }{ Q }

Find A + B + C + D + E + F + G + H + M + N + P + Q 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 A,B,C,E,F,G,H,M,N,P,Q are positive integers not necessarily distinct, C , E C,E are co-prime to each other, A , B A,B are co-prime to each other , G , P G,P are not perfect power of any integer (that it is not a perfect square, cube etc.)

2) Li s ( z ) = k = 1 z k k s \displaystyle { \text{Li} }_{ s }(z)=\sum _{ k=1 }^{ \infty }{ \frac { { z }^{ k } }{ { k }^{ s } } } . It is commonly known as Polylogarithm.

I myself am trying to find the closed form of this integral for about a month, and finally found it.


The answer is 2954.

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3 solutions

Mark Hennings
Jan 16, 2016

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 \pi ): L s 4 ( 1 ) ( π ) = 0 π θ log 2 ( 2 sin 1 2 θ ) d θ = 4 0 1 2 π θ ( log 2 + log ( sin θ ) ) 2 d θ , \mathrm{Ls}_4^{(1)}(\pi) \; = \; -\int_0^\pi \theta \log^2\big(2\sin\tfrac12\theta\big)\,d\theta \; = \; -4\int_0^{\frac12\pi} \theta\big( \log2 + \log(\sin\theta)\big)^2\,d\theta \;, so that I = 0 1 2 π θ log ( sin θ ) d θ = 1 8 π 2 log 2 2 2 log 2 0 1 2 π θ log ( sin θ ) d θ 1 4 L s 4 ( 1 ) ( π ) = 1 8 π 2 log 2 2 2 log 2 { 7 16 ζ ( 3 ) 1 8 π 2 log 2 } 1 4 L s 4 ( 1 ) ( π ) = 1 8 π 2 log 2 2 7 8 log 2 ζ ( 3 ) 1 4 L s 4 ( 1 ) ( π ) = 1 8 π 2 log 2 2 7 8 log 2 ζ ( 3 ) 19 32 ζ ( 4 ) + 1 2 λ 4 ( 1 2 ) \begin{array}{rcl} \displaystyle I \; = \; \int_0^{\frac12\pi} \theta \log(\sin\theta)\,d\theta & = & \displaystyle -\tfrac18\pi^2\log^22 - 2\log 2\int_0^{\frac12\pi} \theta \log(\sin\theta)\,d\theta - \tfrac14\mathrm{Ls}_4^{(1)}(\pi) \\ & = & \tfrac18\pi^2 \log^22 - 2\log 2\left\{ \tfrac{7}{16}\zeta(3) - \tfrac18\pi^2\log2\right\} - \tfrac14\mathrm{Ls}_4^{(1)}(\pi) \\ & = & \tfrac18\pi^2 \log^22 - \tfrac78\log2\zeta(3) - \tfrac14\mathrm{Ls}_4^{(1)}(\pi) \\ & = & \tfrac18\pi^2\log^22 - \tfrac78\log2\zeta(3) - \tfrac{19}{32}\zeta(4) + \tfrac12\lambda_4(\tfrac12) \end{array} where we are using the integral (due to Euler) 0 1 2 π θ log ( sin θ ) d θ = 7 16 ζ ( 3 ) 1 8 π 2 log 2 \int_0^{\frac12\pi} \theta \log(\sin\theta)\,d\theta \; = \; \tfrac{7}{16}\zeta(3) - \tfrac18\pi^2\log2 and we have introduced a special case of the so-called Kummer polylogarithm λ 4 ( 1 2 ) = 2 k = 0 2 1 k ! L i n k ( 1 2 ) log k 2 + 1 4 log 4 2 . \lambda_4(\tfrac12) \; = \; 2\sum_{k=0}^2\tfrac{1}{k!} \mathrm{Li}_{n-k}(\tfrac12) \log^k2 + \tfrac14\log^42 \;. Thus I = 1 8 π 2 log 2 2 7 8 log 2 ζ ( 3 ) 19 32 ζ ( 4 ) + L i 4 ( 1 2 ) + log 2 L i 3 ( 1 2 ) + 1 2 log 2 2 L i 2 ( 1 2 ) + 1 8 log 4 2 = 1 8 π 2 log 2 2 7 8 log 2 ζ ( 3 ) 19 32 ζ ( 4 ) + L i 4 ( 1 2 ) + log 2 { 1 12 π 2 log 2 + 1 6 log 3 2 + 7 8 ζ ( 3 ) } + 1 2 log 2 2 { 1 12 π 2 1 2 log 2 2 } + 1 8 log 4 2 = L i 4 ( 1 2 ) 19 32 ζ ( 4 ) + 1 24 log 4 2 + 1 12 π 2 log 2 2 = L i 4 ( 1 2 ) 19 2880 π 4 + 1 24 log 4 2 + 1 12 π 2 log 2 2 \begin{array}{rcl} I & = & \tfrac18\pi^2\log^22 - \tfrac78\log2\zeta(3) - \tfrac{19}{32}\zeta(4) + \mathrm{Li}_4(\tfrac12) + \log2 \mathrm{Li}_3(\tfrac12) + \tfrac12\log^22 \mathrm{Li}_2(\tfrac12) + \tfrac18\log^42 \\ & = & \tfrac18\pi^2\log^22 - \tfrac78\log2\zeta(3) - \tfrac{19}{32}\zeta(4) + \mathrm{Li}_4(\tfrac12) \\ & & {} + \log2\left\{ -\tfrac{1}{12}\pi^2\log2 + \tfrac16\log^32 + \tfrac78\zeta(3)\right\} + \tfrac12\log^22 \left\{ \tfrac{1}{12}\pi^2 - \tfrac12\log^22\right\} + \tfrac18\log^42 \\ & = & \mathrm{Li}_4(\tfrac12) - \tfrac{19}{32}\zeta(4) + \tfrac{1}{24}\log^42 + \tfrac{1}{12}\pi^2\log^22 \\ & = & \mathrm{Li}_4(\tfrac12) - \tfrac{19}{2880}\pi^4 + \tfrac{1}{24}\log^42 + \tfrac{1}{12}\pi^2\log^22 \end{array} so that the answer is 1 + 2 + 19 + 4 + 2880 + 4 + 2 + 24 + 2 + 2 + 2 + 12 = 2954 1+2+19+4+2880 + 4 + 2 + 24 + 2 + 2 + 2 + 12 \,=\, \boxed{2954} .

( On printer ) +1

Pi Han Goh - 5 years, 4 months ago

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If you are really printing this out, print it again, now I have corrected the typos!

Mark Hennings - 5 years, 4 months ago

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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!

  • Biggest fan.

Pi Han Goh - 5 years, 4 months ago

I have used multiple zeta ... Rest is same evaluating that log sine integral ...

Aman Rajput - 5 years, 4 months ago

Moreover, can you explain how L s 4 ( 1 ) ( π ) Ls_4^{(1)} (\pi) evaluated in terms of λ 4 ( ) \lambda_4() This value is directly given in that... Can you derive please ?

Aman Rajput - 5 years, 4 months ago

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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 ( 1 2 ) 19 8 ζ ( 4 ) ; -\mathrm{Ls}_4^{(1)}(\pi) \; = \; 2\lambda_4(\tfrac12) - \tfrac{19}{8}\zeta(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!

Mark Hennings - 5 years, 4 months ago
Ishan Singh
Jan 14, 2016

Hint: Using the integral representation of H n H_{n} , we have

n = 1 H n t n = ln ( 1 t ) 1 t \displaystyle \sum_{n=1}^{\infty} H_{n} t^n = -\dfrac{\ln (1-t)}{1-t}

Dividing both sides by t t and integrating from 0 0 to x 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 \sin x term in ln 2 ( sin x ) \ln^2 (\sin x ) as e i x e i x 2 i \dfrac{e^{ix}-e^{-ix}}{2i} and take e i x e^{ix} 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 H n n 2 = 2 ζ ( 3 ) \sum_{n=1}^{\infty} \dfrac{H_{n}}{n^2} = 2\zeta (3)

which is left as an exercise for the reader.

This was exactly my intended solution

Ronak Agarwal - 5 years, 5 months ago

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I have a very long solution

Aman Rajput - 5 years, 5 months ago

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Can you post your solution? Thanks.

Pi Han Goh - 5 years, 4 months ago

I don't quite understand what you're saying. Can you explain the "trivial simplications + switching summation and integral" part?

Pi Han Goh - 5 years, 4 months ago

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ln 2 ( sin x ) = [ i x + ln ( 1 e 2 i x ) ln 2 i ] 2 \ln^2 (\sin x) = \left[ ix + \ln(1-e^{-2ix}) - \ln 2i \right]^2 . Expand it and use equation ( ) (*) in the term having ln 2 ( 1 e 2 i x ) \ln^2(1-e^{-2ix}) . 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 ) \ln (1-e^{-2ix}) .

Ishan Singh - 5 years, 4 months ago
Ronak Agarwal
Mar 10, 2015

The answer is = L i 4 ( 1 2 ) 19 π 4 2880 + log 4 ( 2 ) 24 + π 2 log 2 ( 2 ) 12 = { Li }_{ 4 }\left(\dfrac { 1 }{ 2 } \right)-\dfrac { 19{ \pi }^{ 4 } }{ 2880 } +\dfrac { { \log }^{ 4 }(2) }{ 24 } +\dfrac { { \pi }^{ 2 }{ \log }^{ 2 }(2) }{ 12 }

So we have A = 1 , B = 2 , C = 19 , D = 4 , E = 2880 , F = 4 , G = 2 , H = 24 , M = 2 , N = 2 , P = 2 , Q = 12 A=1,B=2,C=19,D=4,E=2880,F=4,G=2,\\ H=24,M=2,N=2,P=2,Q=12

Where's your solution? It's been 3 MONTHS!!

Pi Han Goh - 6 years ago

Where is the solution? : )

Hussein Hijazi - 6 years, 1 month ago

Where is the solution? : )

rajat kharbanda - 6 years, 1 month ago

it's incorrect.... how you evaluate alternating double sum as given in wikipidea

Ciara Sean - 5 years, 7 months ago

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