Recently I have been attempting to find the Laplace Transform of $1-e^{-ax^2}$. I start of with$$\int_0^\infty(1-e^{-ax^2})e^{-nx}dx=\frac{1}{n}-\int_0^\infty e^{-a(x^2+\frac{nx}{a})}dx$$
When I evaluate the integral I am left with $$\int_0^\infty(1-e^{-ax^2})e^{-nx}dx=\frac{1}{n}+e^{\frac{n^2}{4a}}\sqrt{\frac{\pi}{a}}\frac{(erf(\frac{n}{\sqrt{a}})-1)}{2}$$
However, when I sum over the LHS I get $$\int_0^\infty\frac{1-e^{-ax^2}}{e^x-1}dx=\sum_{n=1}^\infty\frac{1}{n}+e^{\frac{n^2}{4a}}\sqrt{\frac{\pi}{a}}\frac{(erf(\frac{n}{\sqrt{a}})-1)}{2}$$
The LHS is a convergent integral while the RHS is a divergent sum. I don't really know where I went wrong, if someone could help me I would be very grateful.