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Does $$I= \int\limits_0^\infty\int\limits_0^x x^{-3/2}e^{y-x}dy\ dx$$ converges, how can I prove it?

I know from Wolfram|Alpha that this integral actually converges to $2\sqrt{\pi}$. This is precisely $$-\Gamma\left(\frac{-1}{2}\right)=-\int\limits_0^\infty x^{-3/2}e^{-x} =2\sqrt{\pi}$$ However when I do $$I=\int\limits_0^\infty x^{-3/2}e^{-x}(e^x-1) dx=-\int\limits_0^\infty x^{-3/2}e^{-x}dx+\int\limits_0^\infty x^{-3/2}dx$$ Which is equivalent to $-\Gamma\left(\frac{-1}{2}\right)+\infty=\infty$.

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    $\begingroup$ Note that $$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x}\,dx$$ only holds when the real part of $z$ is greater than $0$. $\endgroup$ – Brian Moehring Feb 10 at 6:21
  • $\begingroup$ @BrianMoehring I see now, I made some mistakes! $\endgroup$ – Valent Feb 10 at 6:37
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$\int_0^\infty x^{-3/2}e^{-x}\,dx\neq\Gamma(-1/2)$ because the integral diverges. However, $$I=\int_0^\infty x^{-3/2}(1-e^{-x})\,dx=-2x^{-1/2}(1-e^{-x})\Bigg|_0^\infty+2\int_0^\infty x^{-1/2}e^{-x}\,dx=2\Gamma(1/2)=2\sqrt\pi.$$

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You made algebraic mistakes in the last-but-one line. Here is what you can do: $I \leq \int_0^{1} x^{-3/2} (1-e^{-x}) dx+\int_1^{\infty} x^{-3/2}dx <\infty$ since $1-e^{-x} \leq x$ for $0 <x<1$ and $x^{-1/2}$ is integrable on $(0,1)$.

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