Integrals $\int\limits_0^t {{e^u}\log u\operatorname d \!u} $ and $\int\limits_0^t {{e^{ - u}}\log u\operatorname d \!u} $ 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 $t=0$
 A: For $\int_0^te^u\log u\,\mathrm{d}u$,
Consider
\begin{align*}
f(x)
&= \int_0^tu^xe^u\,\mathrm{d}u \\
&= \int_0^tu^x\sum\limits_{n=0}^\infty \frac{u^n}{n!}\,\mathrm{d}u \\
&=\int_0^t\sum\limits_{n=0}^\infty\frac{u^{n+x}}{n!}\,\mathrm{d}u\\
&=\left[\sum\limits_{n=0}^\infty\frac{u^{n+x+1}}{n!\,(n+x+1)}\right]_0^t=\sum\limits_{n=0}^\infty\frac{t^{n+x+1}}{n!\,\left(n+x+1\right)},
\end{align*}
Then
\begin{align*}
f'(x)
&=\int_0^tu^xe^u\log u\,\mathrm{d}u \\
&=\sum\limits_{n=0}^\infty\frac{t^{n+x+1}\log t}{n!\,(n+x+1)}-\sum\limits_{n=0}^\infty\frac{t^{n+x+1}}{n!\,(n+x+1)^2}
\end{align*}
Therefore
\begin{align*}
\int_0^te^u\log u\,\mathrm{d}u
=f'(0)
&=\sum\limits_{n=0}^\infty\frac{t^{n+1}\log t}{n!\,(n+1)}-\sum\limits_{n=0}^\infty\frac{t^{n+1}}{n!\,(n+1)^2} \\
&=\sum\limits_{n=0}^\infty\frac{t^{n+1}\log t}{(n+1)!}-\sum\limits_{n=0}^\infty\frac{t^{n+1}}{(n+1)!\,(n+1)} \\
&=(e^t-1)\log t-\sum\limits_{n=1}^\infty\frac{t^n}{n!\,n}
\end{align*}
For $\int_0^te^{-u}\log u\,\mathrm{d}u$,
Consider
\begin{align*}
g(x)
&= \int_0^tu^xe^{-u}\,\mathrm{d}u \\
&=\int_0^tu^x\sum\limits_{n=0}^\infty\frac{(-1)^nu^n}{n!}\,\mathrm{d}u \\
&=\int_0^t\sum\limits_{n=0}^\infty\frac{(-1)^nu^{n+x}}{n!}\,\mathrm{d}u \\
&=\left[\sum\limits_{n=0}^\infty\frac{(-1)^nu^{n+x+1}}{n!\,(n+x+1)}\right]_0^t=\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+x+1}}{n!\,(n+x+1)}
\end{align*}
Then
\begin{align*}
g'(x)
&=\int_0^tu^xe^{-u}\log u\,\mathrm{d}u \\
&=\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+x+1}\log t}{n!\,(n+x+1)}-\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+x+1}}{n!\,(n+x+1)^2}
\end{align*}
Therefore
\begin{align*}
\int_0^te^{-u}\log u\,\mathrm{d}u=g'(0)
&=\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+1}\log t}{n!\,(n+1)}-\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+1}}{n!\,(n+1)^2} \\
&=\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+1}\log t}{(n+1)!}-\sum\limits_{n=0}^\infty\frac{(-1)^nt^{n+1}}{(n+1)!\,(n+1)} \\
&=(1-e^{-t})\log t+\sum\limits_{n=1}^\infty\frac{(-1)^nt^n}{n!\,n}
\end{align*}
