Question Number 164236 by DAVONG last updated on 15/Jan/22
Commented by DAVONG last updated on 15/Jan/22
$$\mathrm{Helpe}\:\mathrm{pleasd}\:\mathrm{sir} \\ $$
Answered by Kamel last updated on 15/Jan/22
$${I}=\mathrm{2}=\int_{\mathrm{0}} ^{\pi} \int_{\mathrm{0}} ^{{x}} \frac{{sin}\left({x}\right)}{{x}}{dydx} \\ $$
Answered by qaz last updated on 16/Jan/22
![∫_0 ^π ∫_y ^π ((sin x)/x)dxdy =[y∫_y ^π ((sin x)/x)dx]_0 ^π −∫_0 ^π y(−((sin y)/y))dy =∫_0 ^π sin ydy =2](https://www.tinkutara.com/question/Q164382.png)
$$\int_{\mathrm{0}} ^{\pi} \int_{\mathrm{y}} ^{\pi} \frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\mathrm{dxdy} \\ $$$$=\left[\mathrm{y}\int_{\mathrm{y}} ^{\pi} \frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\mathrm{dx}\right]_{\mathrm{0}} ^{\pi} −\int_{\mathrm{0}} ^{\pi} \mathrm{y}\left(−\frac{\mathrm{sin}\:\mathrm{y}}{\mathrm{y}}\right)\mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\:\mathrm{ydy} \\ $$$$=\mathrm{2} \\ $$