1. ## Integral problem

$\displaystyle \int_0^{00}\ {sinx /x} dx\$
the ans is $\displaystyle {pi/2}$

2. i know a solution that covers double integration, but, have you covered double integrals so that you can get my solution?

3. Originally Posted by Krizalid
i know a solution that covers double integration, but, have you covered double integrals so that you can get my solution?
where is ur solution?

4. okay first we integrate by parts to get:

$\displaystyle \int_{0}^{\infty }{\frac{\sin x}{x}\,dx}=\int_{0}^{\infty }{\frac{1-\cos x}{x^{2}}\,dx},$ now on the last integral we use the fact that $\displaystyle \frac1{x^2}=\int_0^\infty te^{-tx}\,dt$ so the last integral becomes $\displaystyle \int_{0}^{\infty }{\int_{0}^{\infty }{\left( te^{-tx}-te^{-tx}\cos x \right)\,dt}\,dx},$ now since $\displaystyle f(x,t)\ge0,$ by using Tonelli, we justify the change of integration order, so this leads to compute $\displaystyle \int_{0}^{\infty }{\left( 1-\frac{t^{2}}{t^{2}+1} \right)\,dt}=\frac{\pi }{2},$ and we are done.