I've been having some problems trying to understand this:
$\int_{0}^{\infty }\frac{e^{-x}}{\sqrt{x}}dx$
I'm required to show that the integral is convergent using the comparison test: If $|f(x)|\le g(x)$ for all $x\ge a$ and $\int_{a}^{b}f(x)dx$ and $\int_{a}^{b}g(x)dx$ exist for every $b\gt a$ then, if $\int_{a}^{b}f(x)dx$ is convergent so is $\int_{a}^{b}g(x)dx$. But how am I supposed to choose $g(x)$ ? This is my wrong work: First I noticed that the integrand is not continuous on $\left[ 0,\infty \right)$, but I deal with that later. Next I break the integral in two parts:
$\int_{0}^{\infty }\frac{e^{-x}}{\sqrt{x}}dx=\int_{0}^{0.1}\frac{e^{-x}}{\sqrt{x}}dx+\int_{0.1}^{\infty }\frac{e^{-x}}{\sqrt{x}}dx$ (I thought the smaller the interval the better.)
I start working on the first integral to test its convergence: By the comparison test, I chose $g(x)=\frac{1}{\sqrt{x}}$, because if $f(x)=\frac{e^{-x}}{\sqrt{x}}$ and $g(x)=\frac{1}{\sqrt{x}}$, then $g(x)\ge f(x)$ on $\left[ 0,0.1 \right]$. Integrating:
$\int_{0}^{0.1}\frac{1}{\sqrt{x}}dx$, since the integrand is not defined at x=0 I write $\int_{0}^{0.1}\frac{1}{\sqrt{x}}dx$ = $\lim_{p \to 0+} \int_{p}^{0.1}\frac{1}{\sqrt{x}}dx$ = $\lim_{p \to 0+} (\frac{2}{\sqrt{10}}-2\sqrt{p})$ = $\frac{2}{\sqrt{10}}$, so I know the first integral is convergent.
Now with $\int_{0.1}^{\infty }\frac{e^{-x}}{\sqrt{x}}dx$ I did almost exactly the same: I choose again $g(x)=\frac{1}{\sqrt{x}}$ because $f(x)=\frac{e^{-x}}{\sqrt{x}}\le g(x)=\frac{1}{\sqrt{x}}$ on $\left[ 0.1,\infty \right)$. But this time I obtain
$\int_{0.1}^{\infty }\frac{1}{\sqrt{x}}dx=\lim_{b \to \infty }\int_{0.1}^{b}\frac{1}{\sqrt{x}}dx=\lim_{b \to \infty }(2\sqrt{b}-\frac{2}{\sqrt{10}})=\infty $, which is divergent. So I got a convergent integral and a divergent integral, something's clearly wrong. I checked the book and in it the integral is broken as
$\int_{0}^{\infty }\frac{e^{-x}}{\sqrt{x}}dx=\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}dx+\int_{1}^{\infty }\frac{e^{-x}}{\sqrt{x}}dx$.
Why $\left[ 0,1 \right]$? Shouldn't any real number on $\left( 0,\infty \right)$ be as valid? What's a good procedure to choose $g(x)$? Thank you.
Assuming $g(x)=\frac{1}{\sqrt{x}}$ and $f(x)=\frac{e^{-x}}{\sqrt{x}}$.
From comparison test for improper integrals:
If $g(x)\geq f(x)\geq 0$ on $[a,\infty)$ then,
In your second case on the interval $[0.1,\infty)$, you have found that $g(x)$ diverges which doesn't imply $f(x)$ diverges.
Now, to find information about $f(x)$ on the whole interval, the motivation of the book is probably:
Let's separate $\int_0^\infty f(x) dx$ such that the divided integrals converge. Say the book divided it to 2 integrals.
$\int_0^\infty f(x) dx$=$\int_0^a f(x) dx$+$\int_a^\infty f(x) dx$
Like how you did , now we analyze $\int_0^a g_1(x) dx$+$\int_a^\infty g_2(x) dx$ instead.
We know that, $\int_0^1 \frac{1}{\sqrt{x}}dx$ converges and thus we have got our first $g(x)$ which satisfies the comparison test condition.
Now, $\int_1^\infty \frac{1}{\sqrt{x}}dx$ diverges so we can't rely on this to be our second $g(x)$. So we search for a new $g(x)$ which satisfies the comparison test.
Let's choose $g(x)=\frac{1}{x^2}$
We know that $\int_1^\infty \frac{1}{x^2}dx$ converges and thus we have got our second $g(x)$ which satisfies the comparison test condition.