Question #9ae3f

1 Answer
Nov 11, 2017

int_0^oo dt/(t^4+1)=(pisqrt(2))/4

Explanation:

I=int_0^oo dt/(t^4+1)

After using t=1/y and dt=-dy/y^2 transformation,

I=int_oo^0 [-dy/y^2]/[(1/y)^4+1]

=int_oo^0 (-y^2*dy)/(y^4+1)

=int_0^oo (y^2*dy)/(y^4+1)

=int_0^oo (t^2*dt)/(t^4+1)

After collecting 2 integrals,

2I=int_0^oo [(t^2+1)*dt]/(t^4+1)

=int_0^oo [(t^2+1)*dt]/[(t^2+sqrt(2)*t+1)(t^2-sqrt(2)*t+1)]

=1/2*int_0^oo [(2t^2+2)*dt]/[(t^2+sqrt(2)*t+1)(t^2-sqrt(2)*t+1)]

=1/2*int_0^oo dt/(t^2+sqrt(2)*t+1)+1/2*int_0^oo dt/(t^2-sqrt(2)*t+1)

=1/2*int_0^oo (2*dt)/(2t^2+2sqrt(2)*t+2)+1/2*int_0^oo (2*dt)/(2t^2-2sqrt(2)*t+2)

=sqrt(2)/2*int_0^oo (sqrt(2)*dt)/[(sqrt(2)*t+1)^2+1]+sqrt(2)/2*int_0^oo (sqrt(2)*dt)/[(sqrt(2)*t-1)^2+1]

=sqrt(2)/2*[Arctan(sqrt(2)*t+1)]_0^oo+sqrt(2)/2*[Arctan(sqrt(2)*t-1)]_0^oo

=(pisqrt(2))/2

Thus, I=(pisqrt(2))/4