Value of convergence of $\displaystyle\int\frac{\sqrt{x}}{1+x^2}$

Question

How to prove that converge $$\int^{\infty}_1\frac{\sqrt{x}}{1+x^2}$$ and find this value.

Answer 1

The answer proves the convergence of the integral.

Here is how you prove the convergence. For large $x$ the integrand behaves as

$$ \frac{\sqrt{x}}{1+x^2} \sim \frac{\sqrt{x}}{x^2}$$

which is an integrable function. Or you can use the inequality

$$ \frac{\sqrt{x}}{1+x^2} \leq \frac{\sqrt{x}}{x^2},\quad x\geq 1. $$

Added:

$$ \int_{1}^{b} x^{-3/2} dx = -2 x^{-1/2}\Big|_{1}^{b}=\dots\, $$

and then take the limit as $b\to \infty$.

Added:

If you are interested in evaluating the integral then start with changing the variables $x=t^2$ which transforms the integral to

$$ I = \int_{1}^{\infty} \frac{2t^2}{1+t^4}dt. $$

Using partial fraction we have

$$I = \int_{1}^{\infty} \frac{1}{t^2+i}dt + \int_{1}^{\infty} \frac{1}{t^2-i}dt $$

To finish the problem you need the result

$$ \int \frac{1}{t^2+a}dt = \frac{1}{\sqrt{a}}\arctan\left(\frac{t}{\sqrt{a}}\right) +C $$

Answer 2

Since the convergence has been established.

For the antiderivative, simplifying the results given by a CAS, the antiderivative is $$\displaystyle\int\frac{\sqrt{x}}{1+x^2}dx=\frac{1}{2 \sqrt{2}}\Big[\log \left(\frac{x+\sqrt{2x} +1}{x-\sqrt{2x}+1}\right)+2 \tan ^{-1}\left(\frac{\sqrt{2x} }{1-x}\right)\Big]$$ and the value of the integral simplifies to $$\int^{\infty}_1\frac{\sqrt{x}}{1+x^2}=\frac{\pi +2 \coth ^{-1}\left(\sqrt{2}\right)}{2 \sqrt{2}}=\frac{\pi +\log \left(3+2 \sqrt{2}\right)}{2 \sqrt{2}}$$ I suppose that for the antiderivative has been used $$\displaystyle\frac{1}{1+x^2}=\frac{1}{2}\Big(\frac{1}{1-ix}+\frac{1}{1+ix}\Big)$$ and that later a change of variable $x=t^2$ has been used.

Working a little more the antiderivative,$$\displaystyle\int\frac{\sqrt{x}}{1+x^2}dx=\frac{1}{\sqrt{2}}\Big[{\cot ^{-1}\left(\frac{1-x}{\sqrt{2x} }\right)-\tanh ^{-1}\left(\frac{1+x}{\sqrt{2x} }\right)}\Big]$$

Answer 3

$$\int^{\infty}_1\frac{\sqrt{x}}{1+x^2}$$ $$=\int^{\infty}_0\frac{\sqrt{x}}{1+x^2} - \int^{1}_0\frac{\sqrt{x}}{1+x^2}$$ The first integral can easily be evaluated by contour integration over a large semicircle on the upper half complex plane. $$\int^{\infty}_0\frac{\sqrt{x}}{1+x^2} = \frac{\pi}{\sqrt{2}} $$

The second integral can be evaluated by a detailed procedure as illustrated