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Question Number 52680 by maxmathsup by imad last updated on 11/Jan/19

$$let{f}_n\left(x\right)=\frac{sin\left(nx\right)}{n^3}andf\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{f}_n\left(x\right)$$$$calculate{\int }_0^{\pi }f\left(x\right)dx.$$

Commented by maxmathsup by imad last updated on 11/Jan/19

$$itsclearthattheserie\mathrm{\Sigma }{f}_n\left(x\right)convergesimpl.andunif.because$$$$\mid f_n\left(x\right)\mid ⩽\frac{1}{n^3}and\mathrm{\Sigma }\frac{1}{n^3}convergeswehave$$$${\int }_0^{\pi }f\left(x\right)dx={\int }_0^{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(nx\right)}{n^3}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}{\int }_0^{\pi }sin\left(nx\right)dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}{[-\frac{1}{n}cos\left(nx\right)]}_0^{\pi }=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}(1-{(-1)}^n)$$$$=2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^4}butwehaveprovedthat\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}=\frac{{\pi }^4}{90}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^4}+\frac{1}{16}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}=\frac{{\pi }^4}{90}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^4}=\frac{{\pi }^4}{90}-\frac{1}{16}\frac{{\pi }^4}{90}$$$$=(1-\frac{1}{16})\frac{{\pi }^4}{90}=\frac{15}{16}\frac{{\pi }^4}{90}=\frac{3.5}{3.3016}{\pi }^4=\frac{{\pi }^4}{6.16}=\frac{{\pi }^4}{96}\Rightarrow$$$${\int }_0^{\pi }f\left(x\right)dx=2.\frac{{\pi }^4}{96}\Rightarrow {\int }_0^{\pi }f\left(x\right)dx=\frac{{\pi }^4}{48}.$$

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