Each of the three following definite integrals are well known to have the same value of $\pi$: $$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x=2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x=\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x=\pi.$$
I like taking the first definite integral as the definition of $\pi$, since it represents half the circumference of the unit circle. The second integral obviously represents the area of the unit circle, but as an exercise I wanted to prove its equality with integral #1 using just elementary integration rules. I was successful once I tried integrating by parts:
$$\begin{align}
\int\sqrt{1-x^2}\,\mathrm{d}x
&=x\sqrt{1-x^2}-\int\frac{-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\sqrt{1-x^2}\,\mathrm{d}x\\
\implies2\int\sqrt{1-x^2}\,\mathrm{d}x&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x\\
\implies2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x&=\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x.
\end{align}$$
Having accomplished this much, I decided I'd like to demonstrate the equality of these two integrals to $\int_{-\infty}^{\infty}\frac{1}{1+x^2}\mathrm{d}x$ as well, in a similarly elementary manner, but I'm stumped as to what to try. Can anyone suggest a substitution or transformation that demonstrates their equality?
Best Answer
These are both immediate using two subsitutions. Everything is even, so split all of them at $0$.
$$\begin{aligned} t=\sqrt{1-x^2}:\quad\int_0^1 \frac{dx}{\sqrt{1-x^2}}= \int_0^1 2\sqrt{1-t^2}\,dt\end{aligned}$$
And:
$$\begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}$$