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How to prove the integral

$$\int_0^{\infty} \int_0^z e^{a(y^2 - z^2) - 2M(y - z)} \; dy \; dz$$

is infinite, where $a$ and $M$ are positive numbers?

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up vote 1 down vote accepted

We have for $z\geq 1$ \begin{align*} \int_0^ze^{a(y^2-z^2)-2M(y-z)}dy&=\int_0^ze^{a((z-t)^2-z^2)+2Mt}dt\\ &=\int_0^ze^{a(-2zt+t^2)+2Mt}dt\\ &\geq \int_0^ze^{-2azt}dt\\ &=-\frac 1{2az}\left[e^{-2atz}\right]_{t=0}^{t=z}\\ &=\frac{1-e^{-2az^2}}{2az}\\ &\geq \frac{1-e^{-2a}}{2az}\geq 0, \end{align*} and $\int_1^{+\infty}\frac 1{z}dz$ is divergent, so we are done.

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The last inequality doesn't hold for all a>0. If we take z=1, 1-e^{-2a}>=e^{-2a} doesn't hold for 0<a<ln2/2. Actually, we can take let it larger than 1/2az. The proof still works. Thanks very much for your help. – Xiangyu Meng Feb 21 '12 at 19:55
You are right, I missed a term, now I think it's correct. – Davide Giraudo Feb 21 '12 at 19:59

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