Evaluate $\int_0^{\infty} \frac{1-e^{-ax}}{x e^x} dx$ I found two different approaches, both is giving the same answer.


*

*Fubini: 
$$
\begin{align}
\int_0^{\infty} \frac{1-e^{-ax}}{x e^x} \,dx &= \int_0^{\infty} e^{-x} \int_0^a e^{-xy} \,dy\, dx \\
&=  \int_0^a \int_0^{\infty} e^{-x(1+y)}\, dx \,dy \\
&= \int_0^{a} \frac{1}{1+y}\, dy\\
&=\log (a+1) , a>-1
\end{align}
$$

*Differentiation of the parameter:
Denote $\displaystyle K(a) = \int_0^{\infty} \frac{1-e^{-ax}}{x e^x}\, dx$, differentiate w.r.t. $a$. Also, note that $K(0)=0$.
$$
\begin{align}
K'(a) &= \int_0^{\infty} \frac{e^{-ax}}{e^x} \,dx\\
&=\int_0^{\infty} e^{-x(a+1)} \,dx\\
&=\frac{1}{a+1}\\
\end{align}
$$
Now we integrate back to get $\displaystyle K(a) = \int K'(a) da = \log(a+1), a>-1$
The requirements of the Fubini theorem are that $f(a,x)$ is a measurable function and $(0,a) \times (0,\infty)$ is a measurable set, right?
To differentiate w.r.t. a parameter, we need that $\displaystyle | e^{-x(a+1)}| \le g(x)$ which has to be an integrable function. Here we could have $g(x)=e^{-x}$ for instance.
So my question now is, whether one of the approaches is more correct than the other. I used 1. in an exam, and got a really low score (so I'm surprised).
 A: I have a third approach relying on interchanging summation and integration:
$$\begin{array}{rcl}
\displaystyle \int_0^{\infty} \frac{1-e^{-ax}}{x e^x} \ \mathrm dx
&=& \displaystyle \int_0^{\infty} \frac{-1}{x e^x} \sum_{n=1}^{\infty} \frac{(-ax)^n}{n!} \ \mathrm dx \\
&=& \displaystyle \int_0^{\infty} \sum_{n=0}^{\infty} (-1)^{n} \frac{a^{n+1}}{(n+1)!} x^n e^{-x} \ \mathrm dx \\
&=& \displaystyle \sum_{n=0}^{\infty} \int_0^{\infty} (-1)^n \frac{a^{n+1}}{(n+1)!} x^n e^{-x} \ \mathrm dx \\
&=& \displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{a^{n+1} n!}{(n+1)!} \\
&=& \displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{a^{n+1}}{n+1} \\
&=& \displaystyle \sum_{n=0}^{\infty} \int_0^a (-x)^n \ \mathrm dx \\
&=& \displaystyle \int_0^a \sum_{n=0}^{\infty} (-x)^n \ \mathrm dx \\
&=& \displaystyle \int_0^a \frac{1}{1-(-x)} \ \mathrm dx \\
&=& \displaystyle \int_0^a \frac{1}{1+x} \ \mathrm dx \\
&=& \displaystyle \log(1+a)
\end{array}$$
provided that $a>-1$.

Which one is more correct?

As long as you have shown that the function in concern is absolutely integrable with a finite integral, all three methods are justified.

I used 1. in an exam, and got a really low score

It is best if you ask your teacher why you got a low score, instead of complaining here. I cannot make a guess for you, as this would be unfair for the teacher.
