$F: \mathbb{R}^+ \to \mathbb{R}, x \mapsto \int_{1}^{+\infty} \frac{\mathrm{d}t}{\sqrt(t) (1+xt)}$

Question

I have to find $\lim \limits_{x\to +\infty} \sqrt(x)F(x)$ ?

First I noticed that $\lim \limits_{x \to +\infty} \frac{1}{\sqrt(t)(1+xt)}=0$. Then I proved that the function is continuous on $\mathbb{R}^+$.

I was wondering if it was a good idea to cut the integral $F$ in two part. Let $R>0$ and $\int_{1}^{+\infty} \frac{\mathrm{d}t}{\sqrt(t) (1+xt)}=\int_{1}^{R} \frac{\mathrm{d}t}{\sqrt(t) (1+xt)}+ \int_{R}^{+\infty} \frac{\mathrm{d}t}{\sqrt(t) (1+xt)}$ because the problem seems to be in the second part.

Thanks in advance !

Answer 1

By making the sustitution $t=u^2$ we get

\begin{align*} \int_1^{\infty}\frac{dt}{\sqrt{t}(1+xt)}&=\int_1^{\infty}\frac{2udu}{u(1+xu^2)}\\ &=\int_1^{\infty}\frac{2du}{1+xu^2}\\ &=\frac{2}{x}\int_1^{\infty}\frac{du}{\frac1x+u^2}\\ &=\frac2x\sqrt{x}\left[\tan^{-1}\left(u\sqrt{x}\right)\right]_1^{\infty}\\ &=\frac2{\sqrt x}\left[\frac{\pi}2-\tan^{-1}(\sqrt x)\right] \end{align*} Then, $$\lim_{x\to\infty}\sqrt xF(x)=\lim_{x\to\infty}\left[\pi-2\tan^{-1}(\sqrt{x})\right]=\pi-2\lim_{x\to\infty}\tan^{-1}(\sqrt{x})=\pi-2\left(\frac{\pi}2\right)=0$$

Answer 2

Let $u=\sqrt{tx}$ then $du=\frac{\sqrt{x}}{2\sqrt{t}}dt$ this integral changes to $\int \frac{2du}{\sqrt{x}(u^2+1)}=\frac{2arctan(u)}{\sqrt{x}}=\frac{2arctan(\sqrt{tx})}{\sqrt{x}}$ substituting the limits of integral for t we have $\frac{\pi-2arctan(\sqrt{x})}{\sqrt{x}}$ thus the limit is $lim_{x\to \infty} (\pi-2arctan(\sqrt{x}))=\pi-2\frac{\pi}{2}=0$

Answer 3

We have $$0<F(x)<\int_1^\infty{dt\over \sqrt{t}\>xt}={1\over x}\int_1^\infty{dt\over t^{3/2}}\qquad(x>0)\ .$$