Find: ∫0π4xln⁡(sin⁡x)dx

Madilyn Fitzgerald

Madilyn Fitzgerald

Answered

2022-01-26

Find: 0π4xln(sinx)dx

Answer & Explanation

Dominique Green

Dominique Green

Expert

2022-01-27Added 11 answers

ln(sinx)=ln2n=1cos(2nx)n
We can use the Fourier series and integrate by parts to obtain
I0π4x[ln(sin(x))]dx=π232ln(2)+14n=11n2[π2sin(π2n)1n(1cos(π2n))]
sin(π2n) is non-zero and alternating for odd n, while cos(π2n) is non-zero and alternating for even n. Therefore,
I=π232ln(2)+π8k=0(1)k(2k+1)214n=11n3132k=1(1)k1k3
The first series is Catalan's constant G, the second one is ζ(3) and the third one is η(3)=34ζ(3) (with the Riemann zeta function ζ and the Dirichlet eta function η), so we obtain
I=π232ln(2)+π8G35128ζ(3)
and your integral is −I.

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