Limit of an integral. $$\lim_{n\to\infty}   \int_{-\infty}^{\infty} \frac{1}{(1+x^2)^n}\,dx $$
Mathematica tells me the answer is 0, but how can I go about actually proving it mathematically?
 A: Simply use this inequality
$$0\leq\int_0^\infty\frac{1}{(1+x^2)^n}\,dx\leq \int_0^\infty\frac{1}{1+nx^2}\,dx=\frac{1}{\sqrt{n}}\int_0^\infty\frac{dt}{1+t^2}=\frac{\pi}{2\sqrt{n}}$$
A: Hint: Apply Dominated Convergence.

Dominated Convergence: Suppose $|f_n(x)|\le g(x)$ where $f_n$ is measurable and $g(x)$ is a non-negative integrable function. If $\lim\limits_{n\to\infty}f_n(x)=f(x)$ ($f$ is the pointwise limit of $f_n\,$), then
  $$
\lim_{n\to\infty}\int f_n\,\mathrm{d}x=\int f\,\mathrm{d}x
$$

A: You can also compute $\displaystyle \int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)^n}$ exactly and argue out what the limit is.
Let $x = \tan(t)$. We then get that
$$I_n = \int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)^n} = \int_{-\pi/2}^{\pi/2} \dfrac{\sec^2(t) dt}{\sec^{2n}(t)} = \int_{-\pi/2}^{\pi/2} \cos^{2n-2}(t) dt = \pi \times \prod_{k=1}^{n-1} \left(1-\dfrac1{2k}\right)$$
Now we have
$$\sum_{k=1}^{\infty} \dfrac1{2k}$$ diverges and hence from here, we have $I_n$ ...
A: Use integration by substitution. Let $x=\tan\theta$. Then
$$ \int_{-\infty}^\infty\frac{1}{(1+x^2)^n}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2(n-1)}\theta d\theta. $$
Now it is not hard to verify 
$$\lim_{n\to\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2(n-1)}\theta d\theta=0.$$
