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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 $t=0$

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Oops, I mistook derivative for antiderivative. Way past my bedtime. –  Henning Makholm Feb 5 '12 at 3:50
    
I suppose you haven't seen this already... –  J. M. Feb 5 '12 at 6:14
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1 Answer

up vote 2 down vote accepted

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*}

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