Question

Question #9ae3f

Answer 1

`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`