Question Number 212139 by mnjuly1970 last updated on 03/Oct/24
$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{I}_{{n}} \:=\:\int_{−\pi} ^{\:\pi} \frac{\:\mathrm{sin}\left({nx}\:\right)}{\left(\mathrm{1}\:+\:{e}^{{x}} \right)\mathrm{sin}{x}}\:{dx}\:=?\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Answered by Frix last updated on 03/Oct/24
![I_n =∫_(−π) ^π ((sin nx)/((e^x +1)sin x))dx =^([t=−x]) =−∫_π ^(−π) ((e^t sin nt)/((e^t +1)sin t))dt =^([t=x]) ∫_(−π) ^π ((e^x sin nx)/((e^x +1)sin x))dx ⇒ 2I_n =∫_(−π) ^π ((sin nx)/((e^x +1)sin x))dx+∫_(−π) ^π ((e^x sin nx)/((e^x +1)sin x))dx= =∫_(−π) ^π ((sin nx)/(sin x))dx ⇒ I_n = { ((0; n=2k+1)),((π; n=2k)) :}](https://www.tinkutara.com/question/Q212144.png)
$${I}_{{n}} =\underset{−\pi} {\overset{\pi} {\int}}\frac{\mathrm{sin}\:{nx}}{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)\mathrm{sin}\:{x}}{dx}\:\overset{\left[{t}=−{x}\right]} {=} \\ $$$$=−\underset{\pi} {\overset{−\pi} {\int}}\frac{\mathrm{e}^{{t}} \mathrm{sin}\:{nt}}{\left(\mathrm{e}^{{t}} +\mathrm{1}\right)\mathrm{sin}\:{t}}{dt}\:\overset{\left[{t}={x}\right]} {=}\underset{−\pi} {\overset{\pi} {\int}}\frac{\mathrm{e}^{{x}} \mathrm{sin}\:{nx}}{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)\mathrm{sin}\:{x}}{dx} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{I}_{{n}} =\underset{−\pi} {\overset{\pi} {\int}}\frac{\mathrm{sin}\:{nx}}{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)\mathrm{sin}\:{x}}{dx}+\underset{−\pi} {\overset{\pi} {\int}}\frac{\mathrm{e}^{{x}} \mathrm{sin}\:{nx}}{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)\mathrm{sin}\:{x}}{dx}= \\ $$$$=\underset{−\pi} {\overset{\pi} {\int}}\frac{\mathrm{sin}\:{nx}}{\mathrm{sin}\:{x}}{dx} \\ $$$$\Rightarrow \\ $$$${I}_{{n}} =\begin{cases}{\mathrm{0};\:{n}=\mathrm{2}{k}+\mathrm{1}}\\{\pi;\:{n}=\mathrm{2}{k}}\end{cases} \\ $$
Commented by mnjuly1970 last updated on 03/Oct/24
$${thanks}\:{a}\:{lot}\:{sir}\:{Frix} \\ $$
Commented by Frix last updated on 03/Oct/24
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}! \\ $$