Question Number 134651 by 0731619177 last updated on 06/Mar/21
Answered by Ar Brandon last updated on 06/Mar/21
![sinx=Σ_(n=0) ^∞ (−1)^n (x^(2n+1) /((2n+1)!)) ⇒sinπx=Σ_(n=0) ^∞ (−1)^n (((πx)^(2n+1) )/((2n+1)!)) I=∫_0 ^∞ (1/(x(1−x^2 )))Σ_(n=0) ^∞ (−1)^n (((πx)^(2n+1) )/((2n+1)!)) =Σ_(n=0) ^∞ (−1)^n (π^(2n+1) /((2n+1)!))∫_0 ^∞ (x^(2n+1) /(x(1−x^2 )))dx =Σ_(n=0) ^∞ (−1)^n (π^(2n+1) /((2n+1)!))∫_0 ^∞ x^(2n) Σ_(k=0) ^∞ x^(2k) =Σ_(n=0) ^∞ (−1)^n (π^(2n+1) /((2n+1)!))Σ_(k=0) ^∞ [(x^(2n+2k+1) /((2n+2k+1)))]_0 ^∞](https://www.tinkutara.com/question/Q134710.png)
$$\mathrm{sinx}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\:\Rightarrow\mathrm{sin}\pi\mathrm{x}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\left(\pi\mathrm{x}\right)^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!} \\ $$$$\mathcal{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\left(\pi\mathrm{x}\right)^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!} \\ $$$$\:\:\:=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\pi^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$\:\:\:=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\pi^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{2n}} \underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{x}^{\mathrm{2k}} \\ $$$$\:\:\:=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\pi^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{2k}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{2k}+\mathrm{1}\right)}\right]_{\mathrm{0}} ^{\infty} \\ $$
Commented by 0731619177 last updated on 06/Mar/21
$${tanks} \\ $$