How do you integrate the following Integral?

#int_(z=0)^oo int_(y=0)^(pi/2) int_(x=0)^pi f(x,y,z) dx dy dz #

with #f(x,y,z)=sin(x+y)ye^(-2z)#

1 Answer
Mar 23, 2018

# (pi-2)/2#

Explanation:

A triple integral has to be calculated step-by-step - one variable at a time:

#int_(z=0)^oo int_(y=0)^(pi/2) int_(x=0)^pi f(x,y,z) dx dy dz equiv int_(z=0)^oo {int_(y=0)^(pi/2) (int_(x=0)^pi f(x,y,z) dx )dy} dz#

We carry out the #x# integral first (here we treat #y# and #z# as constants, at least for the time being)

#int_{x=0}^pi sin(x+y)ye^{-2z}dx = ye^{-2z}int_{x=0}^pi sin(x+y)dx#
#qquad = -ye^{-2z}cos(x+y)|_{x=0}^pi =- ye^{-2z}(cos(pi+y)-cos y)#
#qquad = 2ye^{-2z}cos y#

Next, we carry out the #y# integral

#int_{y=0}^{pi/2}(2ye^{-2z}cos y)dy = 2e^{-2z}int_{y=0}^{pi/2}ycos ydy#

We evaluate this integral by parts

# 2 e^{-2z} (y siny |_{y=0}^{pi/2} - int_{y=0}^{pi/2} sin y dy) #

# qquad = 2e^{-2z} (pi/2+cosy| _{y=0}^{pi/2}) = 2e^{-2z} (pi/2-1)#

Finally, we carry out the final, #z# integral:

# int_{z=0}^oo 2(pi/2-1)e^{-2z}dz = (pi-2) int_{z=0}^oo e^{-2z}dz#
#qquad = -(pi-2)(-1/2e^{-2z})_0^oo=(pi-2)/2#