I would like to calculate $$ F(\alpha)=\int_0^\infty \frac{\log(1+x)}{x}e^{-\alpha x}dx $$ for $\alpha>0.$
Since WolframAlpha provides a complicated expression in terms of hypergeometric functions and the exponential integral $\mathrm{Ei}$ function already for $\alpha=1$, I see little hope for an exact analytical evaluation. Therefore I tried to estimate $F(\alpha)$ at least in some limit, as follows: integrating repeatedly by parts, $$\begin{aligned} f(\alpha)\equiv -F'(\alpha)&=\int_0^\infty \log(1+x)e^{-\alpha x}dx\\ &=-\frac{1}{\alpha}\log(1+x)e^{-\alpha x}\bigg|_0^\infty+\frac{1}{\alpha}\int_0^\infty \frac{e^{-\alpha x}}{1+x}dx\\ &=0-\frac{e^{-\alpha x}}{\alpha^2(1+x)}\bigg|_0^\infty-\frac{1}{\alpha^2}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^2}dx\\ &=\frac{1}{\alpha^2}+\frac{e^{-\alpha x}}{\alpha^3(1+x)^2}\bigg|_0^\infty+\frac{2}{\alpha^3}\int_0^\infty \frac{e^{-\alpha x}}{(1+x)^3}dx\\ &=\frac{1}{\alpha^2}-\frac{1}{\alpha^3}-\frac{2e^{-\alpha x}}{\alpha^4(1+x)^3}\bigg|_0^\infty-\frac{3!}{\alpha^4}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^4}dx. \end{aligned}$$ Proceeding by induction, $$ f(\alpha)=\frac{1}{\alpha^2}\left(1-\frac{1}{\alpha}+\frac{2}{\alpha^2}-\ldots+\frac{(-1)^nn!}{\alpha^n}\right)+(-1)^{n+1}\frac{(n+1)!}{\alpha^{n+2}}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^{n+2}}dx, $$ from which $$\begin{aligned} \lim_{\alpha\to\infty}\alpha^{n+2}\left[f(\alpha)-\sum_{j=0}^{n-1}\frac{(-1)^jj!}{\alpha^{j+2}}\right]&=(-1)^{n}{n!}+(-1)^{n+1}(n+1)!\lim_{\alpha\to\infty}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^{n+2}}dx\\ &=(-1)^{n}{n!} \end{aligned}$$ by dominated convergence. This shows that we have obtained an asymptotic series for $f(\alpha)$ in the limit $\alpha\to\infty$: $$ f(\alpha)\underset{\alpha\to\infty}{\sim}\sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\ . $$ Even if it is of course divergent, this series is Borel summable: $$\boxed{ \sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\overset{B}{=}\int_0^\infty dt\, e^{-t}\sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\frac{t^n}{n!}\,{\color{red}=}\,\frac{1}{\alpha^2}\int_0^\infty e^{-t(1+1/\alpha)}dt=\frac{1}{\alpha(1+\alpha)}}\ . $$ It is therefore tempting to state that $$ f(\alpha)=\int_0^\infty \log(1+x)e^{-\alpha x}dx\overset{?}{=}\frac{1}{\alpha(1+\alpha)} $$ and hence, integrating in $\alpha$ and setting the integration constant to zero because $F(\alpha)\to0$ as $\alpha\to\infty$ by dominated convergence, $$ F(\alpha)=\int_0^\infty \frac{\log(1+x)}{x}e^{-\alpha x}dx \overset{?}{=} \log\frac{1+\alpha}{\alpha}. $$ Strictly speaking, these equalities are wrong, but it appears that they provide reliable numerical approximations to $f(\alpha)$ and $F(\alpha)$ in the $\alpha\to\infty$ limit.
How do I keep track of the degree of approximation involved in the resummation? This was explicit in the asymptotic series, which is however not as nice to write down.
EDIT: Maybe it can be instructive to compare with a similar situation where the exact solution is available. Consider $$ G(\alpha)=\int_0^\infty\frac{\sin x}{x}e^{-\alpha x}dx $$ for $\alpha>0$. Here $$ g(\alpha)\equiv-G'(\alpha)=\int_0^\infty \sin x\, e^{-\alpha x}dx=\mathrm{Im}\left[\int_0^\infty e^{-(\alpha-i)x}dx\right]=\mathrm{Im}\left[\frac{1}{\alpha-i}\right]=\frac{1}{1+\alpha^2} $$ and hence, because $G(\alpha)\to0$ as $\alpha\to\infty$, $$ G(\alpha)=\frac{\pi}{2}-\arctan \alpha. $$ Nevertheless we can work out an asymptotic series for $g(\alpha)$ by integrating by parts. The result is $$ g(\alpha)\underset{\alpha\to\infty}{\sim}\sum_{n=0}^\infty \frac{(-1)^n}{\alpha^{2n+2}}. $$ In this case however the series converges to the exact result $g(\alpha)=1/(1+\alpha^2)$ for $\alpha>1$.
EDIT 2: as was pointed out by Sangchul Lee in the comments, the derivation in the box contains a mistake $$ \sum_{n=0}^{^\infty}\frac{(-1)^n n!}{\alpha^{n+2}}\overset{B}{=}\int_0^\infty \frac{e^{-t}}{\alpha(t+\alpha)}dt\,{\color{red}\neq}\,\frac{1}{\alpha(1+\alpha)}, $$ which is one of the integral representations of $f(\alpha)$ already appearing in the second equation. Therefore, Borel resummation gives the exact result. What I still don't know is how the wrong expressions I derived give a reasonable approximation for large $\alpha$.