Show that the integral from 1 to infinity of 1/(1+x^2) equals the integral from 0 to 1 of 1/(1+t^2) using the substitution t = 1/x

Question

Show that the integral from 1 to infinity of 1/(1+x^2) equals the integral from 0 to 1 of 1/(1+t^2) using the substitution t = 1/x

Answer 1

We are given

\int_{1}^{\infty}\frac{1}{1+x^2} dx.

Now let t = 1/x. Then we have that x = 1/t and dx= -1/t^2 dt. We also have that t = 1 when x = 1 and t = 0 as x goes to positive infinity. All of this now allows us to write the preceding integral as

\int_{1}^{0}\frac{1}{1+(1/t)^{2}}*(-1/t^{2}) dt

= -\int_{1}^{0}\frac{1}{(t^{2}/t^{2})+(1^{2}/t^{2})}*(1/t^{2}) dt

= \int_{0}^{1}\frac{1}{(t^{2} + 1^{2})/t^{2}}*(1/t^{2}) dt

= \int_{0}^{1}\frac{1}{(t^{2} + 1)/t^{2}}*(1/t^{2}) dt

= \int_{0}^{1}\frac{t^{2}}{t^{2} + 1}*(1/t^{2}) dt

= \int_{0}^{1}\frac{1}{t^{2} + 1} dt