Question Number 198141 by Erico last updated on 11/Oct/23
$$\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \:\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\mathrm{dx}=??? \\ $$
Answered by witcher3 last updated on 11/Oct/23
![(1/(sin(πx)))=((2ie^(−iπx) )/(1−e^(−2iπx) )) =(2i)Σ_(n≥0) e^(−iπx(1+2n)) I_n =∫_0 ^1 x(1−x)e^(−iπx(1+2n)) dx =(1/((1+2n)(−iπ)))∫_0 ^1 (1−2x)e^(−iπx(1+2n)) dx =(1/(π^2 (1+2n)^2 ))[(1−2x)e^(−iπx(1+2n)) ]+((2π^2 )/((1+2n)^2 ))∫_0 ^1 e^(−iπ(1+2n)) dx =(2/((1+2n)π^2 ))((2/(iπ(1+2n))))=(4/(iπ^3 (1+2n)^3 )) ∫_0 ^1 ((x(1−x))/(sin(πx)))dx=2iΣ_(n≥0) I_n =(8/π^3 )Σ(1/((1+2n)^3 ))=((7ζ(3))/π^3 )](https://www.tinkutara.com/question/Q198143.png)
$$\frac{\mathrm{1}}{\mathrm{sin}\left(\pi\mathrm{x}\right)}=\frac{\mathrm{2ie}^{−\mathrm{i}\pi\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{2i}\pi\mathrm{x}} }\: \\ $$$$=\left(\mathrm{2i}\right)\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{e}^{−\mathrm{i}\pi\mathrm{x}\left(\mathrm{1}+\mathrm{2n}\right)} \\ $$$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{e}^{−\mathrm{i}\pi\mathrm{x}\left(\mathrm{1}+\mathrm{2n}\right)} \mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2n}\right)\left(−\mathrm{i}\pi\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{2x}\right)\mathrm{e}^{−\mathrm{i}\pi\mathrm{x}\left(\mathrm{1}+\mathrm{2n}\right)} \mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\pi^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2n}\right)^{\mathrm{2}} }\left[\left(\mathrm{1}−\mathrm{2x}\right)\overset{−\mathrm{i}\pi\mathrm{x}\left(\mathrm{1}+\mathrm{2n}\right)} {\mathrm{e}}\right]+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2n}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2n}\right)} \mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{2n}\right)\pi^{\mathrm{2}} }\left(\frac{\mathrm{2}}{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2n}\right)}\right)=\frac{\mathrm{4}}{\mathrm{i}\pi^{\mathrm{3}} \left(\mathrm{1}+\mathrm{2n}\right)^{\mathrm{3}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\mathrm{dx}=\mathrm{2i}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{I}_{\mathrm{n}} \\ $$$$=\frac{\mathrm{8}}{\pi^{\mathrm{3}} }\Sigma\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2n}\right)^{\mathrm{3}} }=\frac{\mathrm{7}\zeta\left(\mathrm{3}\right)}{\pi^{\mathrm{3}} }\: \\ $$$$ \\ $$