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Wolfram MathWorld (citing Abramowitz and Stegun) gives the following result:

$$E_1(x)\equiv\int_x^\infty \frac{e^{-u}}{u}du=-\gamma-\ln x-\sum_{n=1}^\infty\frac{(-1)^nx^n}{n\cdot n!}$$

I am wondering if anybody could supply some details on the proof of this result? In particular, if we Taylor expand $e^{-u}$, we get:

$$E_1(x)=\int_x^\infty \frac{1-u+u^2/2!-u^3/3!+\ldots}{u}du=\left(\ln u+\sum_{n=1}^\infty\frac{(-1)^nu^n}{n\cdot n!}\right)\Bigg|^\infty_x$$

The last two terms of $E_1(x)$ from above (i.e. $-\ln x-\sum_{n=1}^\infty\frac{(-1)^nx^n}{n\cdot n!}$) follow very easily from the lower bound, but I'm not sure how to argue that the upper bound evaluates to the negative Euler-Mascheroni constant $-\gamma$.

Thank you very much for your help!

metamorphy
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JWG
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1 Answers1

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$E_1(x)=f(x)-g(x)$ where $$ f(x)=\int_0^x\frac{1-e^{-t}}{t}\,dt=\int_0^x\sum_{n=1}^\infty\frac{(-t)^{n-1}}{n!}\,dt=-\sum_{n=1}^\infty\frac{(-x)^n}{n\cdot n!}, $$ $$ g(x)=\int_0^x\frac{1-e^{-t}}{t}\,dt-\int_x^\infty\frac{e^{-t}}{t}\,dt=g(1)+\ln x $$ (indeed, write out $g(x)-g(1)$ and simplify), and finally $g(1)=\gamma$ as shown here.

metamorphy
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