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I need to evaluate $$ \int_{0}^{+\infty} \left(\frac{\sqrt\pi}{2}-\int_{0}^{x}\mathrm{e}^{-t^2}dt\right)dx $$ in my homework problem, which should probably be equal to $\frac{1}{2}$. I know $\frac{\sqrt\pi}{2}=\lim\limits_{x\to+\infty}\int_{0}^{x}\mathrm{e}^{-t^2}dt$, but I have no idea on how to evaluate the integral.

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1 Answer 1

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Inasmuch as $\frac{\sqrt \pi}{2}=\int_0^\infty e^{-t^2}\,dt$, we have

$$\begin{align} \int_0^\infty \left( \frac{\sqrt \pi}{2}-\int_0^x e^{-t^2}\,dt\right)\,dx &=\int_0^\infty \int_x^\infty e^{-t^2}\,dt\,dx\\\\ &=\int_0^\infty e^{-t^2}\left(\int_0^t 1\,dx\right) \,dt \end{align}$$

Can you finish?

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    $\begingroup$ Hi Mark ! Nice to see you here today. Cheers :-) $\endgroup$ Feb 13, 2020 at 4:56
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    $\begingroup$ Hi Claude my friend!! How are you? $\endgroup$
    – Mark Viola
    Feb 13, 2020 at 5:22

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