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According to MathWorld, the exponential integral $\operatorname{Ei}(x)=\int^x_{-\infty} \frac{e^t}{t}dt$ can be written as follows:

$$\int^x_{-\infty} \frac{e^t}tdt=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$$

Where $\gamma$ is the Euler-Mascheroni constant.

However, I am curious to know how exactly this was derived.

I was able to show that the derivative of both sides w.r.t. $x$ are the same:

$$\frac d{dx}\int^x_{-\infty} \frac{e^t}{t}dt=\frac d{dx}\left[\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}\right]$$

$$\frac{e^x}x=\frac d{dx}[\gamma]+\frac d{dx}[\ln x]+\sum^\infty_{n=1}\frac d{dx}\left[\frac{x^n}{n\cdot n!}\right]$$

$$\frac{e^x}x=\frac1x+\sum^\infty_{n=1}\frac{nx^{n-1}}{n\cdot n!}$$

$$\frac{e^x}x=\frac1x+\sum^\infty_{n=1}\frac{x^{n-1}}{n!}$$

$$e^x=1+\sum^\infty_{n=1}\frac{x^n}{n!}$$

$$e^x=\sum^\infty_{n=0}\frac{x^n}{n!}$$

Which we know is true by the Maclaurin expansion of $e^x$. However, this only shows that $\int^x_{-\infty} \frac{e^t}tdt$ and $\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$ differ by a constant.

Kyan Cheung
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2 Answers2

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Lets define: $$F(x)=\int_{-\infty}^x\frac{e^t}tdt\tag{1}$$ like you said we can say: $$\frac{e^t}t=\frac1t\sum_{n=0}^\infty\frac{t^n}{n!}=\frac1t+\sum_{n=1}^\infty\frac{t^{n-1}}{n!}$$ and so: $$\int \frac{e^t}tdt=\ln(t)+\sum_{n=1}^\infty\frac{t^n}{n.n!}+C$$

We would like to find a point, $x$, so that we can find the exact solution ourselves and so work out this integration constant. Obviously the problem is the function around $t=0$. If we call our antiderivative $g(t)$ then we know that: $$g(t)=\ln(t)+\sum_{n=1}^\infty\frac{t^n}{n.n!}$$ $$F(x)=g(x)-\lim_{t\to-\infty}g(t)$$

we will have a look at this limit: $$L=\lim_{t\to-\infty}\ln(t)+\sum_{n=1}^\infty\frac{t^n}{n.n!}$$

Now remember that the definition of the Euler-Mascheroni constant is: $$\gamma=\lim_{n\to\infty}\sum_{k=1}^n\frac1k-\ln(n)$$

can you show that $L=-\gamma$?

Henry Lee
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  • Wait. $\ln(t)$ is complex valued for negative $t$ but $\sum_{n=1}^\infty\frac{t^n}{n.n!}$ is real, so wouldn't $L$ be a complex number and not $-\gamma$? Unless I'm missing something. – Kyan Cheung Jan 18 '21 at 13:13
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First of all, this needs a precise formulation (if $x$ is real, then it must be negative for $\int_{-\infty}^x$ to exist, but then $\ln x$ is not defined; if we allow complex values of $x$, then the question of branching arises).

This is why $E_1(z)=\int_z^\infty\frac{e^{-t}}{t}\,dt$ is often considered instead of $\operatorname{Ei}z$ (in case of complex $z$, we choose the branch cut of this function along the negative real axis, as we do for the principal value of $\log z$; this corresponds to the path of integration chosen to stay within lower/upper halfplane according to $z$ itself). Note that for $$f(z)=\int_0^z\frac{1-e^{-t}}{t}\,dt-\int_z^\infty\frac{e^{-t}}{t}\,dt$$ we have $f(z)-f(1)=\log z$, and the equality $f(1)=\gamma$ is the subject of this question. Hence, $$E_1(z)=-\gamma-\log z+\int_0^z\frac{1-e^{-t}}{t}\,dt=-\gamma-\log z-\sum_{n=1}^\infty\frac{(-z)^n}{n\cdot n!}$$ (after termwise integration of the power series for $(1-e^{-t})/t$).

metamorphy
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  • Thanks for your answer. Two questions:

    1. So should I avoid usage of $\operatorname{Ei}(x)$ and use $\operatorname{E}_1(x)$ instead when evaluating integrals?

    2. The question you linked uses the notation $]0,1[$. What exactly does this mean? I've never seen it before.

     – Kyan Cheung Jan 18 '21 at 14:15
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    @Kyky: $\color{red}{1.}$ This is a matter of convenience; I'm basically saying that $\operatorname{Ei}z$ is not defined for positive real $z$ (by the integral above), and for complex $z$, its most natural branch has a cut there. So it looks like $\log(−z)$ rather than $\log z$, as opposed to $E_1(z)$. But nothing prevents you to use $\operatorname{Ei}z=-E_1(-z)$. $\color{red}{2.}$ Well, $]0,1[\ \equiv\ (0,1)$ is just an open interval (I prefer the second version when it doesn't lead to a confusion). – metamorphy Jan 18 '21 at 14:41
  • Appears asked in the past. – metamorphy Feb 11 '21 at 13:23