I am working on an integration, where after applying $e^x = \sum_{n = 0}^\infty\frac{x^n}{n!}$, I have the following transform: \begin{equation} \begin{aligned} \int_0^\infty \log(1+x)e^{-\frac{(b+1)x}{a}}\sum_{m = 0}^\infty\frac{(\frac{x}{2})^{2m}}{m!(m+1)}dx &= \int_0^\infty \log(1+x)e^{-\frac{(b+1)x}{a}}\sum_{m = 0}^\infty\frac{(\frac{x^2}{4})^{m}}{m!}\frac{1}{m+1}dx \\ & = \int_0^\infty \log(1+x)e^{-\frac{(b+1)x}{a}} \frac{4(e^{\frac{x^2}{4}}-1)}{x^2}dx\\ \end{aligned} \end{equation}
I found myself helplessness with this integration while trying with integration by parts method.
Would anyone please guide me on this integral?
Thank you in advance!
This integral really depends on a single parameter $c=\frac{b+1}{a}$ as can be seen:
$$ \int_0^\infty \log(1+x)e^{\frac{ax^2 - 4(b+1)x}{4a}}dx=\int_0^\infty \log(1+x)e^{x^2/4 - c x}dx $$
However, no matter which value $c$ takes, the integral diverges because of the $e^{x^2}$ term which goes to infinity as $x \to \infty$. (And $x^2$ grows faster than $cx$ for any $c$).
$$ \int_0^\infty \log(1+x)e^{\frac{ax^2 - 4(b+1)x}{4a}}dx=\infty $$
Update:
In answer to the questions about the sum in the edited post:
$$\sum_{m = 0}^\infty\frac{y^m}{m!(m+1)}= \frac{1}{y} \sum_{m = 0}^\infty\frac{y^{m+1}}{m!(m+1)}=$$ $$=\frac{1}{y} \sum_{m = 0}^\infty\frac{y^{m+1}}{(m+1)!}=\frac{1}{y} \sum_{m = 1}^\infty\frac{y^m}{m!}=$$ $$=\frac{1}{y} \left(e^y-1 \right)$$ Now substitute $y=x^2/4$.
As this expression still leads to a diverging integral (now with additional problem at $x=0$), there must me some error in the prior reasoning which led to the sum in the first place.