Help evaluating $\int _0^{+\infty}\frac{1-e^{-\lambda x}}{xe^x}dx$

Question

I need help evaluating $$F(\lambda) = \int _0^{+\infty}\frac{1-e^{-\lambda x}}{xe^x}dx$$ $\lambda \gt 1$

I know that I should use the derivate under the sign of the integral but I'm having trouble how to show that this function holds the properties to do so.

To be able to use the derivate under the integral the function $F(\lambda)= \int _0^{+\infty} f(x,\lambda)dx$ should hold these properties

$1.$ The function $f(x,\lambda)$ and it's partial derivative $f'_\lambda$ should be continous for $0 \le x \lt +\infty$ and $\lambda \gt -1$

For this to be true I took $$f(x, \lambda)= \begin{cases} \frac{1-e^{-\lambda x}}{xe^x} , & x \neq0 \\[2ex] \lambda & x = 0 \end{cases}$$

$2.$ The integral $ \int _0^{+\infty} f(x,\lambda)dx$ is convergent for $\lambda \gt -1$

$3.$ The integral $ \int _0^{+\infty} f'_\lambda(x,\lambda)dx$ should be uniformly convergent for $\lambda \gt -1$

The second point is where I'm having trouble because I don't know how to prove that the integral converges

Answer 1

Hint:

The integrand is bounded in $[0,1]$, hence you can consider convergence of

$$\int_1^\infty\dfrac{e^{-x}}xdx-\int_1^\infty\dfrac{e^{(-1-\lambda)x}}xdx.$$

$$\int_1^\infty\dfrac{e^{-ax}}xdx<\int_1^\infty e^{-ax}dx=\frac{e^{-a}}a.$$

Answer 2

You can calculate the integral directly as follows using Fubini-Tonelli (FT)

\begin{eqnarray*} \int _0^{+\infty}\frac{1-e^{-\lambda x}}{xe^x}dx & = & \int _{x=0}^{+\infty}e^{-x}\int_{y=0}^{\lambda}e^{-yx}\;dy\;dx \\ & \stackrel{FT}{=} & \int_{y=0}^{\lambda}\int_{x=0}^{+\infty}e^{-(1+y)x}\;dx\;dy \\ & = & \int_{y=0}^{\lambda} \frac{1}{1+y}\; dy \\ & = & \ln(1+\lambda) \end{eqnarray*}

Note, that $e^{-(1+y)x}$ is non-negative and measurable on $\mathbb{R}\times [0,\lambda]$. So, according to Fubini-Tonelli the order of integration can be exchanged without changing the value of the integral (see the link above).