Is it possible to get series expansion at $x = 0$ of $F : x \mapsto \int_0^{\alpha} \frac{e^{-\frac{t}{x}}}{1 + t} \textrm{d}t$? Let be $\alpha > 0$. Let be $f : (x, t) \mapsto \dfrac{e^{-\frac{t}{x}}}{1 + t}$, $F : x \mapsto \int_0^{\alpha} f(x, t) \textrm{d}t$.
I am struggling to establish a series expansion of $F$ at $x = 0$ due to $e^{-\frac{t}{x}}$ being delicate to study.
I tried:


*

*Dominated convergence theorem, I showed that $\dfrac{F(x)}{x} \to 1$ when $x \to 0$.

*Series expansion of $f(x, t)$ and performing an interversion, but got a Laurent series (unsure of the validity of the operations).

*Change of variable, but make appear a sort of improper integral.


Let be $n \in \mathbb{N}$, is there a way to find $(a_n)$ a sequence of reals such that:
$\begin{equation*}
F(x) = \sum\limits_{k=0}^{n} a_k x^k + o(x^n)
\end{equation*}$
More generally, how to proceed when there is a term such as $\exp(-t h(x))$ which is complex to analyze when $h(x) \to +\infty$ ?
 A: Here some integrations by parts are just fine to obtain a series expansion as $x \to 0^+$, 
$$
\begin{align}
& \int_0^{\alpha} \frac{e^{-\frac{t}{x}}}{1 + t}\: dt
\\\\&=\left[ -xe^{-t/x}\:\frac{1}{1 + t}\right]_0^\alpha
-x\int^\alpha_0 e^{-t/x}\frac{1}{(1 + t)^2}\,dt
\\\\&=\color{red}{x}-xe^{-\alpha/x}-x\left(\left[ -xe^{-t/x}\:\frac{1}{(1 + t)^2}\right]_0^\alpha
-2x\int^\alpha_0 e^{-t/x}\frac{1}{(1 + t)^3}\,dt\right)
\\\\&=\color{red}{x-x^2}-xe^{-\alpha/x}-x^2e^{-\alpha/x}+2x^2\int^\alpha_0 e^{-t/x}\frac{1}{(1 + t)^3}\,dt
\\\\&=\color{red}{x-x^2}+o(x^2)
\end{align}
$$ and so on.
A: As Olivier Oloa did, integration by parts leads to the following steps
$$\int_0^{\alpha} \frac{e^{-\frac{t}{x}}}{1 + t}\, dt=x-\frac{x e^{-\frac{\alpha }{x}}}{\alpha +1}-x \int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^2}\,dt$$
$$\int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^2}\,dt=x-\frac{x e^{-\frac{\alpha }{x}}}{(\alpha
   +1)^2}-2x\int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^3}$$
$$\int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^3}\,dt=x-\frac{x e^{-\frac{\alpha }{x}}}{(\alpha
   +1)^3}-3x\int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^4}\,dt$$
$$\int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^4}\,dt=x-\frac{x e^{-\frac{\alpha }{x}}}{(\alpha
   +1)^4}-4x\int_0^{\alpha}\frac{ e^{-\frac{t}{x}}}{(1+t)^5}\,dt$$ and so on.
Since $\alpha >0$ and $x\to 0$, the second terms become negligible and the expansion becomes something like
$$x-x^2+2 x^3-6 x^4+24 x^5+O\left(x^6\right)$$
For sure, the same result would be obtained using
$$\int_0^{\alpha} \frac{e^{-\frac{t}{x}}}{1 + t}\, dt=e^{\frac{1}{x}} \left(\Gamma \left(0,\frac{1}{x}\right)-\Gamma \left(0,\frac{\alpha
   +1}{x}\right)\right)$$ where appear the incomplete gamma function. 
Using the asymptotics for large values of $y$
$$\Gamma(0,y)=e^{-y}
   \left(\frac{1}{y}-\frac{1}{y^2}+\frac{2}{y^3}-\frac{6}{y^4}+\frac{2
   4}{y^5}+O\left(\frac{1}{y^6}\right)\right)$$ would lead to 
$$\int_0^{\alpha} \frac{e^{-\frac{t}{x}}}{1 + t}\, dt=\left(x-x^2+2 x^3-6 x^4+24
   x^5+O\left(x^6\right)\right)-e^{-\frac{\alpha }{x}} \left(\frac{x}{\alpha +1}-\frac{x^2}{(\alpha +1)^2}+\frac{2
   x^3}{(\alpha +1)^3}-\frac{6 x^4}{(\alpha +1)^4}+\frac{24 x^5}{(\alpha
   +1)^5}+O\left(x^6\right)\right)$$
For a numerical check, using $\alpha=1$ and $x=\frac 1{10}$, the above approximation would lead to $\frac{2291}{25000}-\frac{1193}{25000 e^{10}}\approx 0.0916378$ while the "exact" value would be $\approx 0.0915612$. Using only the $x^k$ terms, the result wold be $\frac{2291}{25000}=0.09164$.
