For what value does the following improper integral exist Find $m$ such that $\displaystyle\int_{-\infty}^\infty \frac{1}{(1+x^2)^m} \, dx$ is finite.
I tried to substitute $x$ with $\tan\theta$ but got stuck.
 A: $(1+x^2)^m \sim x^{2m} as~ x \to \infty$, this property may help, I think.
A: There is usually more than one way to skin an integral, and your original approach is certainly tenable.  We present an approach here that uses that way forward.
If we enforce the substitution $x=\tan(\theta)$, then we have
$$\begin{align}
\int_{-\infty}^\infty \frac{1}{(1+x^2)^m}\,dx&=2\int_{0}^{\pi/2}\cos^{2(m-1)}(\theta)\,d\theta\\\\
&=2\int_0^{\pi/2}\sin^{2(m-1)}(\theta)\,d\theta \tag 1
\end{align}$$
Obviously, the integral on the right-hand side of $(1)$ is integrable for $m\ge 1$. But what happens if $m<1$?  
For $m<1$, the exponent on the sine function is negative.  Inasmuch as the sine function vanishes for $\theta=0$, the integral is improper.  However, if we recall from elementary geometry that for $0\le \theta\le \pi/2$ the sine function satisfies the inequalities
$$\frac2\pi \,\theta \le \sin(\theta)\le \theta \tag 2$$
then, using $(2)$, we have for any $0<\epsilon <\pi/2$ and $m<1$
$$\int_{\epsilon}^{\pi/2}\frac{1}{\left(\theta\right)^{2(1-m)}}\,d\theta\le \int_0^{\pi/2}\sin^{2(m-1)}(\theta)\,d\theta \le \frac\pi 2\,\int_{\epsilon}^{\pi/2}\frac{1}{\left(\theta\right)^{2(1-m)}}\,d\theta\tag 3$$
Noting that 
$$\lim_{\epsilon\to 0^+}\int_{\epsilon}^{\pi/2}\frac{1}{\left(\theta\right)^{2(1-m)}}\,d\theta=\begin{cases}\frac{(\pi/2)^{2m-1}}{2m-1}&,m>1/2\\\\\infty&,m\le 1/2\end{cases}$$ 
Therefore, the integral of interest converges for $m>1/2$ and diverges for $m\le 1/2$.
