Evaluating $\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx $

Question

I am working on this improper integral,

$$\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx$$

First, I separate the integral into two pieces, I have

$$\int_{0}^{\infty}\frac{e^{-x^2}}{x}dx -\int_{0}^{\infty}\frac{e^{-x}}{x}dx=I_1+I_2$$

I know that $I_2$ can use double integral,

$$ I_2=\int_0^\infty \int _1^\infty e^{-tx} dt dx = \int _1^\infty \frac{1}{t}dt $$

$$ \int_0^\infty (\frac{e^{-tx}}{-x}) \Bigg|_{1}^\infty dx = \int_1^\infty \frac{1}{t}dt =\ln t \Bigg|_{1}^{\infty}=\infty$$

But I don't know how to do $I_1$, can someone give me a hit or suggestion?

Thanks!

$\frac{e^{-x^2}}{x}=\frac{e^{-x^2}}{x^2}x$, then substitute $u=x^2$ to get $\frac{1}{2}I_2$. — vadim123, Feb 23 '15 at 02:48
Well, sadly $I_1$ also does not converge (it goes to $-\infty$). Now the actual answer to the first integral is $\frac{\gamma}{2}$ where $\gamma$ is Euler's constant, also known as the Euler-Mascheroni constant ($\approx 0.577216$), but to be honest I don't know how to derive this result. — Uncountable, Feb 23 '15 at 02:50

score 1 · Answer 1 · answered Feb 23 '15 at 02:48

1

hint: $\dfrac{e^{-x^2}}{x} = \dfrac{1}{x^2}\cdot xe^{-x^2}$. Let $u = x^2$

answered Feb 23 '15 at 02:48

DeepSea

78,689

kobe · Answer 2 · 2015-03-23T15:20:52.827

Some care must be taken when trying evaluating this integral. Fist consider the partial integrals

$$\int_a^b \frac{e^{-x^2} - e^{-x}}{x}\, dx \quad (b > a > 0).$$

We have

$$\int_a^b \frac{e^{-x^2} - e^{-x}}{x}\, dx = \int_a^b \frac{e^{-x^2}}{x}\, dx - \int_a^b \frac{e^{-x}}{x}\, dx = \frac{1}{2}\int_{a^2}^{b^2} \frac{e^{-u}}{u}\, du - \int_a^b \frac{e^{-x}}{x}\, dx,$$

using the substitution $u = x^2$. By integration by parts,

$$\int_a^b \frac{e^{-x}}{x}\, dx = e^{-b}\ln b - e^{-a}\ln a + \int_a^b e^{-x}\ln x\, dx$$

and

$$\frac{1}{2}\int_{a^2}^{b^2} \frac{e^{-u}}{u}\, du = e^{-b^2}\ln b - e^{-a^2}\ln a + \frac{1}{2}\int_{a^2}^{b^2} e^{-x}\ln x\, dx.$$

Therefore

\begin{align}&\int_a^b \frac{e^{-x^2} - e^{-x}}{x}\, dx\\ & = (e^{-b^2} - e^{-b})\ln b - (e^{-a^2} - e^{-a})\ln a + \frac{1}{2}\int_{a^2}^{b^2} e^{-x}\ln x\, dx - \int_a^b e^{-x}\ln x\, dx.\\ \end{align}

Since $(e^{-x^2} - e^{-x})\ln x$ tends to $0$ as $x\to 0^+$ and as $x\to \infty$, taking the limit as $a \to 0^+$ and $b\to \infty$ results in

$$\int_0^\infty \frac{e^{-x^2} - e^{-x}}{x}\, dx = -\frac{1}{2}\int_0^\infty e^{-x}\ln x\, dx = \frac{1}{2}\gamma,$$

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

I don't understand the second step where you convert one integral to-> a^2 to b^2 ... You've substituted u=x^2 .. so it should be $\sqrt{u}$ in the denominator..and 'du' instead of 'dx' right? — Panglossian Oporopolist, Mar 23 '15 at 15:16
I had a typo there - $dx$ should be $du$. Otherwise the integral is correct. With $u = x^2$, $du = 2x, dx$, thus $du/(2u) = dx/x$. — kobe, Mar 23 '15 at 15:20

Evaluating $\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx $

2 Answers2