I would like to integrate the following with respect to x.
y*e^x
--------
x^2
I can't find the right substitution.
Thanks!
$\displaystyle y\int\frac{e^{x}}{x^{2}}dx$.
This is not easily done with elementary methods. That is why you can not find the right substitution.
It can be represented by the exponential integral, Ei.
Which is represented by $\displaystyle Ei(a,t)=\int_{1}^{\infty}e^{-tz}t^{-a}dt$
Yours is:
$\displaystyle \frac{-e^{x}}{x}-Ei(1,-x)$
Thanks for this but I should have added that it is not to be done with exponential integrals, or I don't think so because it has never been and will never be discussed in the course I'm taking.
I was wondering about some kind of amazing trick. There must be one. Our problems are always done so that there is an amazing trick.
Thanks although.
One approach I see quickly is just use series.
$\displaystyle \frac{e^{x}}{x^{2}}=\frac{1}{2}+\frac{1}{x}+\frac{ 1}{x^{2}}+\frac{1}{6}x+\frac{1}{24}x^{2}+\frac{1}{ 120}x^{3}+\frac{1}{720}x^{4}+................$
This in closed form is $\displaystyle \sum_{k=1}^{\infty}\frac{x^{k-1}}{(k+1)!}+\frac{1}{x}+\frac{1}{x^{2}}$
If we integrate, we get:
$\displaystyle ln(x)-\frac{1}{x}+\frac{x}{2}+\frac{x^{2}}{12}+\frac{x^{ 3}}{72}+\frac{x^{4}}{480}+....$
$\displaystyle =\sum_{k=1}^{\infty}\frac{x^{k}}{k(k+1)!}-\frac{1}{x}+ln(x)$