Csdrhrt

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

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

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

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.)

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:

$int_{[0, infty) times[0, infty)} e^{-(x^2+y^2)} dx dy = int_0^{infty} big(int_0^{fracpi2}re^{-r^2}dtheta)dr = fracpi2int_0^{infty}re^{-r^2}dr = fracpi2 frac12 = fracpi4$

Therefore:

$int_{[0, infty)} e ^ {-x^2}dx = big(fracpi4big)^{frac12}$

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.

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

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

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

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

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

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

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

P.D.: Excuse my English, please.