Integration of Exponential

Question

I am trying to integrate this function $f(x)=e^{-c/x}$.

$$\int_{a}^b e^{-c/x} dx \\$$

where $c$ is just a constant and $0<a<b$. But $u$ subsititution leads to me to an integration by parts which leads to another integration by parts that keeps going. Is there a closed form solution to this integral?? I appreciate any help. Thank you.

Update:

I derived this function from a conditional pdf. $f(y|x)=x^{-1}e^{-y/x}$ and $f(x)=x$ on $[0,\sqrt{2}]$. By using Bayes Theorem you get a joint pdf $f(x,y)=e^{-y/x}$. Now using the joint pdf I am trying to solve for partial pdf $f_y(y)$. In the problem above I generalize it to $[a,b]$ and $y=c$.

Answer 1

Considering $$I=\int e^{-\frac c x} dx $$ make a change of variable $-\frac c x=y$ that is to say $x=-\frac c y$, $dx=\frac c {y^2}dy$ to make $$I=c\int \frac{ e^y}{y^2}dy$$ Now, integration by parts $$u=e^y\quad du=e^y dy\quad dv=\frac{ dy}{y^2}\quad v=-\frac 1 y$$This gives $$I=c\left(-\frac {e^y} y+\int \frac{ e^y}{y}dy\right)=c \left(\text{Ei}(y)-\frac{e^y}{y}\right)$$ where appears the definition of the exponential integral function.

There is no way to express the result in terms of elementary functions.

However, expanding the exponential integral, for real $y$, you have $$\text{Ei}(y)=\gamma +\log(|y|)+\sum_{k=1}^\infty\frac {y^k}{k\, k!}$$