$\displaystyle \int_0^{00}\ {sinx /x} dx\$

the ans is $\displaystyle {pi/2}$

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- Feb 12th 2010, 05:12 AMchialin4Integral problem
$\displaystyle \int_0^{00}\ {sinx /x} dx\$

the ans is $\displaystyle {pi/2}$ - Feb 12th 2010, 05:23 AMKrizalid
i know a solution that covers double integration, but, have you covered double integrals so that you can get my solution?

- Feb 12th 2010, 05:48 AMchialin4
- Feb 12th 2010, 06:39 AMKrizalid
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.