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

Answer 1

∫(1/(1+x^2)dx from 1 to ∞ = ....

Substitute t=1/x

Then x=1/t

dx = -1/t^2 dt [ derivative dx/dt = -1/t^2 by Power Rule applied to x = t^(-1) ]

x=1 gives t=1

x —> ∞ gives t=0

Note:

1+x^2 = 1 + 1/t^2 = (t^2+1)/t^2

So

1/(1+x^2) = t^2/(t^2+1)

Finally

∫(1/(1+x^2)dx from 1 to ∞ = ∫[t^2/(t^2+1)] [-1/t^2 dt] from 1 to 0

= — ∫[1/(t^2+1)] dt from 1 to 0

= + ∫[1/(t^2+1)] dt from 0 to 1 (the desired result)