Ok, I want to find
$\int\limits_0^t {{e^u}\log udu} $ and $\int\limits_0^t {{e^{ - u}}\log udu} $
I'm thinking as follows
$d\left( {{e^u}\log u} \right) = {e^u}\log udu + \frac{{{e^u}}}{u}du$
$d\left( {{e^{ - u}}\log u} \right) = - {e^{ - u}}\log udu + \frac{{{e^{ - u}}}}{u}du$
Thus I put
$\int {{e^u}\log udu} = {e^u}\log u - Ei\left( u \right)$
$\int {{e^{ - u}}\log udu} = Ei\left( { - u} \right) - {e^{ - u}}\log u$
But then I want to integrate over $(0,t)$. My principal concern is proving that for $t\to 0$
$ Ei\left( { - t} \right) - {e^{ - u}}\log u\to \gamma$
${e^u}\log u - Ei\left( t \right) \to -\gamma$
How would you solve this?
EDIT: Ok I've found that $Ei(t) = \gamma + \log(t) + z + \frac{z^2}{4} + \cdots + \frac{z^n}{n n!}+\cdots$
The new question would be: How do I prove such expansion?
I mean, I know that
$\int \frac{e^x}{x} dx = \log x + \sum_{n>0} \frac{x^n}{n n!}$
But then again where does $\gamma$ appear? Since $Ei(t) = \int_{-\infty}^t \frac{e^u}{u} du$ I'd need to find the limit for $x \to -\infty$, which should be $\gamma$.
Is guess you could also use $Ei(t) = \log t + \int_0^t {\frac{e^u-1}{u}du}$ which is a more "natural" definition in the sense the integral is always converging for finite $t$ and the logarithm instantly discloses the discontinuity at $t=0$