Finiteness of Improper integral

Question

How do I show that if $t> 1/2$, then

$$\int_R \dfrac{1}{(1+x^2)^t}dx $$ is finite?

I tried to use the p-test, but since $\int_R \dfrac{1}{x^p}$ diverges for all $p$, I could not show the finiteness of above integral

Answer 1

HINT:

For $t>0$ and $x\ne 0$, we have

$$(1+x^2)^t>x^{2t}\implies \frac{1}{(1+x^2)^t}\le \frac{1}{x^{2t}}$$

And we have $$\int_0^L \frac{1}{(1+x^2)^t}\,dt=\int_0^1 \frac{1}{(1+x^2)^t}\,dt+\int_1^L \frac{1}{(1+x^2)^t}\,dt$$

Answer 2

As soon as $t\geq 0$ the integral $\int_{-1}^{1}\frac{dx}{(1+x^2)^t}$ is bounded by the constant $2$, so there are no integrability issues close to the origin. And as soon as $t>\frac{1}{2}$ we have $$ 0\leq \int_{-\infty}^{-1}\frac{dx}{(1+x^2)^t}=\int_{1}^{+\infty}\frac{dx}{(1+x^2)^t}\leq \int_{1}^{+\infty}\frac{dx}{x^{2t}}=\frac{1}{2t-1}.$$ The exact value of the integral is given by $\frac{\Gamma\left(t-\frac{1}{2}\right)}{\Gamma(t)}\sqrt{\pi}$ due to Euler's beta function.
For moderately large values of $t$ we have $\int_{\mathbb{R}}\frac{dx}{(1+x^2)^t}\approx\sqrt{\frac{\pi}{t}}\left(1-\frac{3}{8t}\right)^{-1}.$

Answer 3

$\displaystyle\Re\left(-2t + 1 < 0\right) \implies \bbox[10px,#ffe,border:1px dotted navy]{\displaystyle\Re\left(t\right)\ >\ {1 \over 2}}$.