Computing an improper integral with respect to a parameter

Question

I am motivated by this problem.Let us compute an improper integral with respect to a parameter:$$F(x)=\int_{1}^{\infty}\frac{e^{-xy}-1}{y^{3}}dy,\quad x\in[0,\infty).$$

The following is my effort:

$$F(x)=\int_{1}^{\infty}\frac{e^{-xy}-1}{y^{3}}dy=(-\frac{1}{2})\int_{1}^{\infty}(e^{-xy}-1)d(y^{-2})=\\(-\frac{1}{2})[y^{-2}(e^{-xy}-1)\vert_{1}^{\infty}-\int_{1}^{\infty}y^{-2}d(e^{-xy}-1)]=(-\frac{1}{2})[(1-e^{-x})+x\int_{1}^{\infty}\frac{e^{-xy}}{y^2}dy] .\quad (1)$$

$$\text{since}\quad F'(x)=\int_{1}^{\infty}f'_{x}(x,y)dy,\quad F'(x)=\int_{1}^{\infty}\frac{-e^{-xy}}{y^{2}}dy.\quad (2)$$ $(1)+(2)\Longrightarrow F(x)=\frac{e^{-x}-1}{2}+\frac{x}{2}F'(x)$

For every $x>0$,we have $F(x)=x^2(C+\int\frac{1-e^{-x}}{x^3}dx).$ If we can find the constant $C$ ,then my question will be well done. The indefinite Integral $\int\frac{1-e^{-x}}{x^3}dx$ is not easy to deal with .How can I compute the value of $C$?

In another way,if you have some good ideas about how to solve this question ,please give me some hints.Any help is going to be appreciated!

Answer 1

Your solution is correct. For finding $C$ we need initial value such as $F(0)=1$ and $F(1)=2$ for example by substitution in solution $C$ will be calculated simply.

Answer 2

This is not an answer to the question but it is far too long for a comment.

Considering the problem of the antiderivative $$I=\int\frac{e^{-xy}-1}{y^{3}}\,dy$$ let us make a first change of variable $y=\frac tx$ which makes $$I=x^2\int \frac{\left(e^{-t}-1\right) }{t^3} \,dt=x^2\,\Big(\int\frac{e^{-t}}{t^3}\,dt-\int\frac{1}{t^3}\,dt\Big)$$ and the problem is left with the first antiderivative which makes thinking about the exponential integral function $$\int\frac{e^{-t}}{t}\,dt=\text{Ei}(-t)$$ After some manipulations $$J=\int\frac{e^{-t}}{t^3}\,dt=\frac{\text{Ei}(-t)}{2}-\frac{e^{-t}}{2 t^2}+\frac{e^{-t}}{2 t}$$ and then $$I=\frac{x^2} 2\,\Big(\text{Ei}(-t)-\frac{e^{-t}}{t^2}+\frac{e^{-t}}{t}+\frac{1}{t^2} \Big)$$ Going back to to $y$ does not make any problem and the only thing left is to use the bounds for integration. Using again special functions, $$F(x)=\int_{1}^{\infty}\frac{e^{-xy}-1}{y^{3}}dy=\frac{x^2 }{2} \text{Ei}(-x)-\frac{1}{2} e^{-x} (x-1)-\frac{1}{2}$$ (provided that $\Re(x)>0 $) where appear the hyperbolic sine and cosine integrals.

Developed as a series at $x=0$, one could get $$F(x)=-x-\frac{x^2}{4} (2 \log (x)+2 \gamma -3)+\frac{x^3}{6}-\frac{x^4}{48}+O\left(x^5\right)$$

So, back to the original problem, around $x=0$, $$\frac{F(x)}{\log(1+x)}=-1-\frac{x}{4} (2 \log (x)+2 \gamma -1)-\frac{x^2}{8} (2 \log (x)+2 \gamma -5)+O\left(x^3\right)$$

Edit

In case you are concerned, if $$I_n=\int \frac{e^x}{x^n}\,dx$$ ($n>1)$, we can establish a nice and simple recurrence relation $$I_n=\frac 1 {n-1}\Big(I_{n-1}-\frac {e^x}{x^{n-1}}\Big)$$ with $I_1=\text{Ei}(x)$.