This came up while I was looking at the asymptotic behavior of $f(x)=\int_0^x \frac{e^t-1}{t}dt=\sum_{h=1}^\infty\frac{x^h}{h\cdot h!}$, which is a nice and entire function, as $x\to -\infty$, namely that I think $f(x)\sim -\ln(-x)-\gamma$ as $x\to -\infty$. I've reduced that to the problem of proving $$\int_{-1}^0 \frac{e^x+e^{1/x}-1}{x}dx=\gamma$$ where $\gamma\approx 0.577$ is the Euler-Mascheroni constant.
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$\begingroup$ Are you familiar with the exponential integral? $\endgroup$– teadawg1337Jan 21, 2017 at 17:52
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$\begingroup$ @Dr.MV That's true if $x$ is positive, but what about $\displaystyle\lim_{x\to 0^-}e^{1/x}$? $\endgroup$– teadawg1337Jan 21, 2017 at 17:55
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1$\begingroup$ Mathematica shows that result, so it's converge. $\endgroup$– NosratiJan 21, 2017 at 17:57
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2 Answers
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First, enforce the substitution $x\to -1/x$, we obtain
$$\int_{-1}^0\frac{e^{1/x}}{x}\,dx=-\int_1^\infty \frac{e^{-x}}{x}\,dx \tag 1$$
Second, integrating by parts the integral on the right-hand side of $(1)$ with $u=e^{-x}$ and $v=\log(x)$ reveals
$$\begin{align} \int_{-1}^0\frac{e^{1/x}}{x}\,dx&=-\int_1^\infty e^{-x}\log(x)\,dx\\\\ &=-\int_0^\infty e^{-x}\log(x)\,dx+\int_0^1 e^{-x}\log(x)\,dx\\\\ &=\gamma +\int_0^1 e^{-x}\log(x)\,dx\tag 2 \end{align}$$
Finally, integrating by parts the integral on the right-hand side of $(2)$ with $u=\log(x)$ and $v=-(e^{-x}-1)$ reveals
$$\begin{align} \int_{-1}^0\frac{e^{1/x}}{x}\,dx&=\gamma +\int_0^1 \frac{e^{-x}-1}{x}\,dx\\\\ &=\gamma -\int_{-1}^0\frac{e^x-1}{x}\,dx \tag 3 \end{align}$$
whence solving $(3)$ for $\gamma$ yields the coveted relationship
$$\bbox[5px,border:2px solid #C0A000]{\gamma=\int_{-1}^0 \frac{e^x+e^{1/x}-1}{x}\,dx}$$
And we are done!
At the end of THIS ANSWER, I showed that $\gamma$ as given by $-\int_0^\infty \log(x)e^{-x}\,dx=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n \frac1k\right)$, the latter of which is often given as the definition of $\gamma$.
answered Jan 21, 2017 at 18:29
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$\begingroup$ @myglasses Thank you! Much appreciative. $\endgroup$ Jan 21, 2017 at 19:55
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$\begingroup$ You beat me to the punch... I became distracted in the midst of writing up my answer, but I've posted it anyway as a second approach. Your proof is better than mine, though! $\endgroup$ Jan 21, 2017 at 20:12
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1$\begingroup$ How did you find $\int_0^\infty e^{-x}\ln(x)dx=-\gamma$? By using the power series of $e^{-x}$, switching the sum and integral and integrating termwise I got $\int_0^z e^{-x}\ln(x)dx=(1-e^{-z})\ln(z)+f(-z)$ which means the asymptotic statement I'm after and this integral are equivalent. $\endgroup$– SophieJan 21, 2017 at 20:46
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2$\begingroup$ Sophie. How do you define $\gamma$? If one uses the limit definition, $\gamma=\lim_{n\to \infty}\left(H_n-\log(n)\right)$,then have a look at the following. $$\int_0^\infty \log(x) e^{-x}\,dx=\lim_{n\to \infty}\int_0^n \left(1-\frac xn \right)^n\,\log(x)\,dx$$Now, let $u=1-x/n$, expand the logarithm in a Taylor series, integrate term by term and you will recover the limit definition of $\gamma$. $\endgroup$ Jan 21, 2017 at 20:57
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I'm going to take a different approach than Dr. MV, and evaluate the integral directly. First, due to the singularity at $x=0$, the integral should be interpreted as $\displaystyle \lim_{\epsilon\to 0^-}\int_{-1}^\epsilon \frac{e^x+e^{1/x}-1}x dx$. With this in mind, we can integrate. $$\int\frac{e^x+e^{1/x}-1}x dx=\mathrm{Ei}(x)-\mathrm{Ei}\left(\frac1x\right)-\ln{x}+C\tag{1}$$ It's easy to see that the value at $x=-1$ is $-i\pi$, but the value near $0$ is a tad more complex (no pun intended). Luckily, the first term can be quickly disposed of: $$-\lim_{x\to 0^-}\mathrm{Ei}\left(\frac1x\right)=-\lim_{t\to\infty}\mathrm{Ei}(-t)=0.$$ This leaves us with $\displaystyle\lim_{x\to 0^-}\left[\mathrm{Ei}(x)-\ln{x}\right]$. For $x<0$, we can use the series expansion for $E_1(z)$ to write $$\mathrm{Ei}(x)=\gamma+\ln(-x)+\sum_{n=1}^\infty\frac{x^n}{n\,n!}.\tag{2}$$ We now have $$\lim_{x\to 0^-}\left[\mathrm{Ei}(x)-\ln{x}\right]=\lim_{x\to 0^-}\left[\gamma+\ln(-x)-\ln{x}+\sum_{n=1}^\infty\frac{x^n}{n\,n!}\right]=\gamma-i\pi\tag{3}$$ and may conclude that $$\int_{-1}^0\frac{e^x+e^{1/x}-1}x dx=\gamma+i\pi-i\pi=\gamma.\tag{4}$$ Q.E.D.
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$\begingroup$ This works. My resistance to appealing to the Exponential Integral and its expansion ia that one might ask, "How does one arrive at the leading term?" But (+1) for the solution. $\endgroup$ Jan 21, 2017 at 20:26
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$\begingroup$ @Dr.MV Fair enough. From my limited understanding, it appears to have something to do with the derivative of the Gamma function. In fact, the expansion for $\displaystyle\int\frac{\mathrm{Ei}(x)}{x}dx$ includes $\displaystyle\frac{\partial^2}{\partial x^2}\left[\frac{\Gamma(x)}2\right]_{x=1}$ $\endgroup$ Jan 21, 2017 at 20:42