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Question Number 162001 by SANOGO last updated on 25/Dec/21

$$natureof:$$$${\int }_0^{+oo}\frac{sint}{e^t-1}dt$$

Answered by mathmax by abdo last updated on 25/Dec/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{{\mathrm{e}}^{\mathrm{t}}-1}\mathrm{dt}\Rightarrow \mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}}{1-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nt}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}\mathrm{sint}\mathrm{dt}$$$$\mathrm{but}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}\mathrm{sint}\mathrm{dt}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}+\mathrm{it}}\mathrm{dt}\right)$$$$\mathrm{and}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-(\mathrm{n}+1)+\mathrm{i})\mathrm{t}}\mathrm{dt}={\left[\frac{1}{-(\mathrm{n}+1)+\mathrm{i}}{\mathrm{e}}^{(-(\mathrm{n}+1)+\mathrm{i})\mathrm{t}}\right]}_0^{\mathrm{\infty }}$$$$=-\frac{1}{\mathrm{n}+1-\mathrm{i}}(-1)=\frac{1}{\mathrm{n}+1-\mathrm{i}}=\frac{\mathrm{n}+1+\mathrm{i}}{{(\mathrm{n}+1)}^2+1}\Rightarrow$$$$\mathrm{\Psi }=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2+1}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2+1}\mathrm{et}\mathrm{cette}\mathrm{serie}\mathrm{est}\mathrm{convergente}$$$$\mathrm{donc}\mathrm{\Psi }\mathrm{est}\mathrm{cv}.$$

Commented by Ar Brandon last updated on 25/Dec/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+{\varphi }^2}=\frac{\pi \coth \left(\pi \varphi \right)}{2\varphi }-\frac{1}{2{\varphi }^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}=\frac{1}{2}(\pi \coth \left(\pi \right)-1)$$

Commented by SANOGO last updated on 25/Dec/21

$$mercibien$$

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