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.


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 $y=c$.

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    $\begingroup$ I believe that (for $c\not=0$) this integral does not have a closed form in terms of elementary functions. $\endgroup$ Nov 22, 2015 at 3:02
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    $\begingroup$ This will end up being an exponential integral - just a special function, but not anything expressible in terms of elementary functions. $\endgroup$
    – Ron Gordon
    Nov 22, 2015 at 3:03
  • $\begingroup$ Thanks! I thought so too. But leads to me a problem. I made an update above. Thank you for your help! $\endgroup$
    – jessica
    Nov 22, 2015 at 3:24
  • $\begingroup$ @jessica Just in case you are not aware: By an elementary function we mean a function that is made of logarithm, exponential, or trigonometric or inverse trigonometric functions in terms of operations of addition, subtraction, multiplication, division, and composition. $\endgroup$
    – Yes
    Nov 22, 2015 at 3:34
  • $\begingroup$ Thanks! Yes, I understood. Is the way I am attempting to derive the partial pdf for y correct? $\endgroup$
    – jessica
    Nov 22, 2015 at 4:00

1 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 $$\text{Ei}(y)=\gamma +\log(|y|)+\sum_{k=1}^\infty\frac {y^k}{k\, k!}$$


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