GATE 2019:EC

Question

The value of integral $\int_{0}^{\pi }\int_{y}^{\pi }\frac{\sin x}{x}dxdy$ is equal to_________

Answer 1

$P=\int_{y}^{\pi }\frac{\sin x}{x}dx=\frac{1}{2}\left [ -\frac{1}{x} \cos x+\log x.\sin x\right ]$

                               $=\frac{1}{2}\left [ \frac{1}{\pi }+\frac{1}{y}\cos y-\log y \sin y\right ]$

 

 

Now, separately find $\int_{0}^{\pi }\frac{\cos y}{y} dy=\frac{\sin y}{y}+\int \log y \sin y .dx............................(1)$

 

$\int_{0}^{\pi }P.dx$ $=\frac{1}{2}\left [ \int \frac{1}{\pi }+\int \frac{1}{y}\cos y-\int \log y \sin y\right ] dy$

putting eq.$(1)$ here we got

                  $=\frac{1}{2}\left [ \int \frac{1}{\pi }+\frac{\sin y}{y}+\int \log y \sin y-\int \log y \sin y\right ] dy$

                   $=\frac{1}{2}\left [ \frac{y}{\pi }+\frac{\sin y}{y} \right ]_{0}^{\pi }=\frac{1}{2}=0.5$

Comments

This solution is wrong.