Does the improper integral $\int_0^\infty e^{-x^2}dx$ converge?

Question

I want to show the convergence of the following improper integral $\int_0^\infty e^{-x^2}dx$. I try to use comparison test for integrals $x≥0$, $-x ≥0$, $-x^2≥0$ then $e^{-x^2}≤1$. So am ending with the fact that $\int_0^\infty e^{-x^2}dx$ converges if $\int_0^\infty dx$ converges but I don’t appreciate this. Thanks

Answer 1

Write

$$\int_0^\infty e^{-x^2} \, dx = \int_0^1 e^{-x^2} \, dx + \int_1^\infty e^{-x^2} \, dx$$

The first integral on the right hand side is of a bounded function over a bounded interval, hence finite. For the second notice that we have $x^2 > x$ when $x > 1$. Therefore, $e^{-x^2} < e^{-x}$ for $x > 1$ and it follows that

$$\int_1^\infty e^{-x^2} \, dx < \int_1^\infty e^{-x} \, dx$$ $$= \lim_{x \to \infty} -e^{-x} + e^{-1} = 1/e < \infty$$

Answer 2

I'm assuming you're asking about the convergence of $\int_0^\infty e^{-x^2}dx$. The easiest way that I can think of to prove this is to note that $e^{-x^2}$ is continuous and bounded, and hence integrable, on the interval $[0,1]$, and that on the remaining unbounded interval $[1,\infty)$ it is a function everywhere bounded in absolute value by a function $e^{-x}$ that is integrable on that interval (seeing that $\int_1^\infty e^{-x}dx$ converges is a simple calculation, since $e^{-x}$ has an easy antiderivative). Thus, by the comparison test, $\int_0^\infty e^{-x^2}dx$ converges. I think this is the idea you're referring to in your question.

$\int_0^\infty e^{-x^2}dx$ does not exist if $\int_0^\infty dx$ does, since the latter one most certainly does not converge (it equals $\lim_{x \to \infty} x = \infty$). While it is true that $e^{-x^2} \le 1$ on $[0,\infty)$, this fact isn't really helpful, because the comparison test only gives us information when the bounding function is itself integrable (otherwise you get absurdities like claiming that $\int_0^\infty e^{-x}dx$ diverges because $e^{-x} \le x $ and $\int_0^\infty x dx$ diverges.)

Answer 3

It does, and you can also compute its value:

$\int_{[0, \infty) \times[0, \infty)} e^{-(x^2+y^2)} dx dy = \int_{[0, \infty)}\big(\int_{[0, \infty)} e^{-(x^2+y^2)} dy \big)dx = \int_{[0, \infty)} e ^ {-x^2}\big(\int_{[0, \infty)} e^{-y^2} dy \big)dx =$ $= \int_{[0, \infty)} e ^ {-x^2}dx \ \ \int_{[0, \infty)} e^{-y^2} dy = \big( \int_{[0, \infty)} e ^ {-x^2}dx \big)^2$

Then we can use polar coordinates:

$\int_{[0, \infty) \times[0, \infty)} e^{-(x^2+y^2)} dx dy = \int_0^{\infty} \big(\int_0^{\frac\pi2}re^{-r^2}d\theta)dr = \frac\pi2\int_0^{\infty}re^{-r^2}dr = \frac\pi2 \frac12 = \frac\pi4$

Therefore:

$\int_{[0, \infty)} e ^ {-x^2}dx = \big(\frac\pi4\big)^{\frac12}$

Answer 4

Exercise 3-41 of Spivak's "Calculus on Manifolds" outlines how to prove that $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$ using polar coordinates, a related double integral, and two useful families of regions over which to evaluate the double integral. The description is actually quite long. Ironically, he ends the problem with a quote attributed to Lord Kelvin (William Thomson), who was trying to convey the meaning of "mathematician" to a class: "A mathematician is one to whom "that" (i.e., $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$) is as obvious as that twice two makes four is to you." A bit harsh, I daresay, on Lord Kelvin's part. The reality being conveyed by Spivak is that it's obvious once you've done the necessary work.

Answer 5

Let $$f(x) = \frac{1}{{e^x}^{2}}$$ Now let $$g(x) = \frac{1}{x^2 + 1}$$ $$\lim _{x\rightarrow \infty} = 0 $$ and $g(x)$ is a convergent integral, so $f(x)$ is convergent too.

Answer 6

Denote $$f(x):=\left(\int_0^{x} e^{-t^2}{\rm d}t\right)^2,~~~g(x):=\int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2}{\rm d}t.$$ Then $$f'(x)=2\int_0^{x} e^{-t^2}{\rm d}t\left(\int_0^{x} e^{-t^2}{\rm d}t\right)'=2e^{-x^2}\int_0^{x} e^{-t^2}{\rm d}t,$$ and $$g'(x)=\int_0^1 \dfrac{\partial \left(\dfrac{e^{-x^2(1+t^2)}}{1+t^2}\right)}{\partial x}{\rm d}t=\int_0^1 \dfrac{-2x(1+t^2)e^{-x^2(1+t^2)}}{1+t^2}{\rm d}t=-2xe^{-x^2}\int_0^1 e^{-x^2t^2}{\rm d}t.$$ Let $u:=tx$. Then $x=\dfrac{u}{t}$ and ${\rm d}t=\dfrac{{\rm d}u}{x}.$ We obtain $$g'(x)=-2xe^{-x^2}\int_0^x \frac{e^{-u^2}}{x}{\rm d}u=-2e^{-x^2}\int_0^x e^{-u^2}{\rm d}u=-f'(x),$$ which implies $$f'(x)+g'(x)=0.$$ Therefore $$f(x)+g(x)\equiv C,$$ where $C$ is some constant. Thus $$C=f(0)+g(0)=0+\int_0^1 \frac{1}{1+t^2}{\rm d}t=[\arctan t]_0^1=\frac{\pi}{4}.$$ Hence $$f(+\infty)+g(+\infty)=\frac{\pi}{4}.$$ But we can derive $g(+\infty)=0$, and it follows that $$f(+\infty)=\frac{\pi}{4},$$ which gives that $$\int_0^{+\infty} e^{-t^2}{\rm d}t=\sqrt{\frac{\pi}{4}}=\frac{\sqrt{\pi}}{2}.$$

Answer 7

A simple answer as @neemy… Remember the function of Normal probability

$$N(\mu,\sigma)=\displaystyle{\int_{-\infty}^{+\infty}{\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\;dx}=\bf{1}}$$

For $\mu=0$ and $\sigma=1$ $$\displaystyle{\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\;dx}=\bf{1}}$$

Introduce a change of variable with its respective Jacobian $$x=\sqrt{2}\;u\\dx=\sqrt{2}\;du$$

and result $$\displaystyle{\int_{-\infty}^{+\infty}{\frac{\sqrt{2}}{\sqrt{2\pi}}e^{-u^2}\;du}=\bf{1}} \Rightarrow \displaystyle{\int_{-\infty}^{+\infty}{e^{-u^2}\;du}=\bf{\sqrt{\pi}}}$$

The Normal is even function, therefore $$\displaystyle{\int_0^{+\infty}{e^{-u^2}\;du}=\frac{\sqrt{\pi}}{2}}$$

P.D.: Excuse my English, please.