I'm stuck trying to workout this integral $$\int_0^\infty dx \; \log (1-\exp(-a x^2)) $$ for $a>0$. So, I tried first to do Taylor Series of the integrand, that is $$\begin{align} \int_0^\infty dx \; \log (1-\exp(-a x^2)) &= \int_0^\infty dx \sum_{k=1}^\infty \frac{(-1)^{k}(-\exp(-a x^2))^k}{k}\\
&= \sum_{k+1}^\infty \frac{(-1)^{k+1}}{k} \int_0^\infty dx \exp(-akx^2)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac{\pi^{1/2}}{2 (ak)^{1/2}}\\
&= \left( \frac{\pi}{4a} \right)^{1/2} \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{3/2}} \end{align} $$
This last series, clearly converges. I also tried to do something in the lines of using Leibniz rule for integration, but nothing promising yielded.
Any help is appreciated! Thanks in advance.
Best Answer
Found one mistake! Here I lay my corrected computation $$\begin{align} \int_0^\infty dx \; \log (1-\exp(-a x^2)) &= \int_0^\infty dx \sum_{k=1}^\infty \frac{(-1)^{k}(-\exp(-a x^2))^k}{k}\\ &= \sum_{k+1}^\infty \frac{1}{k} \int_0^\infty dx \exp(-akx^2)= \left( \frac{\pi}{4a} \right)^{1/2} \sum_{k=1}^\infty \frac{1}{k^{3/2}} \\ &=\left( \frac{\pi}{4a} \right)^2 \zeta(3/2) \end{align} $$ Now it yields my expected result!