Hot Integral - 22

Calculus Level 5

∫ 0 π t 3 log ⁡ 8 ( 2 sin ⁡ t ) d t \displaystyle \int\limits_{0}^{\pi} t^3\log^8(2\sin t) \, dt

If the above integral can be expressed as : A π C B + D π F ζ ( G ) H E + I π J ζ ( K ) L + M π N ζ ( O ) ζ ( P ) + Q U π R ζ ( S ) ζ ( T ) \frac{A\pi^C}{B}+\frac{D\pi^F\zeta(G)^H}{E}+I\pi^J\zeta(K)^L + M\pi^N\zeta(O)\zeta(P)+\frac{Q}{U}\pi^R\zeta(S)\zeta(T)

Evaluate A + B + C + D + E + F + G + H + I + J + K + L + M + N + O + P + Q + R + S + T + U A+B+C+D+E+F+G+H+I+J+K+L+M+N+O+P+Q+R+S+T+U

Details and Assumptions

  • A , B , . . . , S , T , U A,B,...,S,T,U all are integers.

  • gcd ⁡ ( A , B ) = gcd ⁡ ( D , E ) = gcd ⁡ ( Q , U ) = 1 \gcd(A,B)=\gcd(D,E)=\gcd(Q,U)=1

This is a part of Hot Integrals


The answer is 153380.

Aman Rajput by Aman Rajput , 25, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Jan 19, 2016

If we define F ( ν , a ) = 2 ν ∫ 0 π sin ⁡ ν t sin ⁡ a t d t = π sin ⁡ ( 1 2 a π ) ( ν + 1 ) B ( 1 2 ( ν + a + 2 ) , 1 2 ( ν − a + 2 ) ) F(\nu,a) \; = \; 2^\nu\int_0^\pi \sin^\nu t \sin at\,dt \; = \; \frac{\pi \sin\big(\frac12a\pi\big)}{(\nu+1) B\big(\frac12(\nu+a+2),\frac12(\nu-a+2)\big)} then the desired integral is I = − ∂ 11 F ∂ ν 8 ∂ a 3 ( 0 , 0 ) . I \; = \; -\frac{\partial^{11}F}{\partial \nu^8 \partial a^3}(0,0) \;. Now G ( ν ) = − ∂ 3 F ∂ a 3 ( ν , 0 ) = π 2 Γ ( ν + 2 ) ( π 2 + 6 ψ ′ ( 1 + 1 2 ν ) ) 8 ( ν + 1 ) Γ ( 1 2 ν + 1 ) 2 = π 2 Γ ( ν + 1 ) ( π 2 + 6 ψ ′ ( 1 + 1 2 ν ) ) 8 Γ ( 1 2 ν + 1 ) 2 \begin{array}{rcl} G(\nu) & = & \displaystyle-\frac{\partial^3F}{\partial a^3}(\nu,0) \; = \; \frac{\pi^2\Gamma(\nu+2)\big(\pi^2 + 6\psi'(1 + \tfrac12\nu)\big)}{8(\nu+1)\Gamma(\tfrac12\nu + 1)^2} \\ & = &\displaystyle \frac{\pi^2\Gamma(\nu+1)\big(\pi^2 + 6\psi'(1 + \tfrac12\nu)\big)}{8\Gamma(\tfrac12\nu + 1)^2}\end{array} After a whole bundle of simplification (!), we have I = G ( 8 ) ( 0 ) = 39223 101376 π 12 + 735 16 π 6 ζ ( 3 ) 2 + 2835 π 2 ζ ( 5 ) 2 + 5670 π 2 ζ ( 3 ) ζ ( 7 ) + 3465 4 π 4 ζ ( 3 ) ζ ( 5 ) \begin{array}{rcl} I & = & G^{(8)}(0) \\ & = & \tfrac{39223}{101376}\pi^{12} + \tfrac{735}{16}\pi^6 \zeta(3)^2 + 2835\pi^2 \zeta(5)^2 + 5670\pi^2\zeta(3)\zeta(7) + \tfrac{3465}{4}\pi^4 \zeta(3)\zeta(5) \end{array} making the answer equal to 39223 + 101376 + 12 + 735 + 16 + 6 + 3 + 2 + 2835 + 2 + 5 + 2 + 5670 + 2 + 3 + 7 + 3465 + 4 + 3 + 5 + 4 = 153380 \begin{array}{l}39223 + 101376 + 12 + 735 + 16 + 6 + 3 + 2 + 2835 + 2 + 5 + 2 \\ +5670 + 2 + 3 + 7 + 3465 + 4 + 3 + 5 + 4 \; = \; \boxed{153380} \end{array}

11 Helpful 0 Interesting 1 Brilliant 0 Confused

can you explain the first line in the proof. it is too scary.

Srikanth Tupurani - 1 year, 9 months ago

Log in to reply

If you integrate z a − 1 ( z − z − 1 ) ν z^{a-1}\big(z - z^{-1}\big)^\nu around the semicircular contour formed by the region of the circle ∣ z ∣ ≤ 1 |z| \le 1 for R e z ≥ 0 \mathfrak{Re}\,z \ge 0 , indented at 0 , 1 , − 1 0,1,-1 suitably, then we obtain this result for ν > − 1 \nu > -1 and a > ν a > \nu . It is then possible to use analytic continuation to show that the result holds for all ν > − 1 \nu > -1 and for all a a . This is a simple variation of an integral known to Euler (integrating cos ⁡ a θ cos ⁡ ν θ \cos a\theta \cos^\nu\theta from − 1 2 π -\tfrac12\pi to 1 2 π \tfrac12\pi ). I guess that Euler probably knew this one as well!

Mark Hennings - 1 year, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...