<s>[INST] Proof existence of $\int_0^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx$ - is this proof correct? I have to show that the integral
$$ \int_0^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx $$
exists.
My approach: Let $b > 1$, then the integral
$$ \int_1^b \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx $$
exists because
$$
0 < \int_1^b \frac{1}{\sqrt x} \underbrace{\frac{1}{1+x^2}}_{\leq \frac{1}{x^2}} \,\mathrm dx \leq \int_1^b \frac{1}{x^{5/2}} \,\mathrm dx = \left.-\frac{2}{3} x^{-3/2} \right|_1^b = \underbrace{-\frac{2}{3} b^{-3/2}}_{\xrightarrow{b \to \infty}{0}} + \frac{2}{3} \xrightarrow{b \to \infty} \frac{2}{3}.
$$
Let $0 < \epsilon < 1$, then
$$ \int_\epsilon^1 \frac{1}{\sqrt x} \underbrace{\frac{1}{1+x^2}}_{<1} \,\mathrm dx < \int_\epsilon^1 \frac{1}{\sqrt x} \,\mathrm dx = \left.2 \sqrt x \right|_\epsilon^1 = 2(1 - \sqrt \epsilon) \xrightarrow{\epsilon \searrow 0} 2 $$
and in total
$$ 0 < \int_0^\infty \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx = \underbrace{\int_0^1 \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx}_{\leq 2} + \underbrace{\int_1^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx}_{\leq \frac{2}{3}} \leq 2 + \frac{2}{3}. $$
Is this a proper proof?
 [/INST] Looks fine, here is another approach :
\begin{aligned}&\bullet \ f:x\mapsto\frac{1}{\sqrt{x}\left(1+x\right)}\textrm{ is continuous on }\left(0,+\infty\right)\cdot\\ &\bullet \ \lim_{x\to 0}{\frac{x^{\frac{3}{4}}}{\sqrt{x}\left(1+x\right)}}=0\textrm{, and thus }f\left(x\right)=\underset{\overset{x\to 0}{}}{\mathcal{o}}\left(x^{-\frac{3}{4}}\right)\textrm{, and }x\mapsto x^{-\frac{3}{4}}\textrm{ is integrable on }\left(0,a\right]\textrm{ for}\\ &\textrm{any }a\in\left(0,+\infty\right) \cdot\\ &\bullet \ \lim_{x\to +\infty}{\frac{x^{\frac{3}{2}}}{\sqrt{x}\left(1+x\right)}}=0\textrm{, thus }f\left(x\right)=\underset{\overset{x\to +\infty}{}}{\mathcal{o}}\left(x^{-\frac{3}{2}}\right)\textrm{, and }x\mapsto x^{-\frac{3}{2}}\textrm{ is integrable on }\left[a,+\infty\right)\\ &\textrm{for any }a\in\left(0,+\infty\right) \cdot \end{aligned}
Thus, our $ f $ is integrable on $ \left(0,+\infty\right)\cdot $
</s>