#intint_D cosxy dxdy#, where #D: {[x>=0],[y>=0],[x>=y]}#?

1 Answer
Ultrilliam
Jun 25, 2018

This integration is in the first quadrant, under the line #y = x#

#intint_D cosxy \ dx \dy#

#=color(blue)( int_(x=0)^oo dx \ int_(y=0)^(x) dy qquad cos xy ) color(green)(equiv) color(red)( int_(y = 0) ^oo dy \ int_(x=y)^(oo) dx qquad cos xy)#

Following the blue integration:

#= int_(x=0)^oo dx qquad [ (sin xy)/x]_(y=0)^x#

#= int_(x=0)^oo dx qquad (sin x^2)/x #

#z = x^2 qquad dz = 2 x \ dx = 2 sqrtz \ dx#

#= int_0^oo dz 1/(2sqrtz) qquad (sin z)/sqrtz #

#=1/2 underbrace( int_0^oo dz qquad (sin z)/z)_("Feynmann Integration") = 1/2 * pi/2 = pi/4 #

For that last integration, a sketch:

Create function:

  • #I(a ) = int_0^oo dz qquad (sin z)/z e^(- a z) qquad triangle qquad a>=0#

Differentiate wrt #a#:

  • #I'(a ) = int_0^oo dz qquad - z (sin z)/z \ e^(- a z)#

#=- int_0^oo dz qquad sin z \ e^(- a z)#

That, after 2 rounds of IBP, is:

#I'(a) =- 1/(a^2 + 1)#

Integrate wrt #a#

#I(a) = - arctan (a) + C qquad square#

Combining:

  • #triangle: qquad lim_(a to oo) int_0^oo dz qquad (sin z)/z e^(- a z) = 0#

  • #square: qquad I(a to oo) = - arctan (a to oo) + C = 0#

#implies C = pi/2#

#implies I(a) = - arctan (a) + pi/2#

And:

  • #I(0) = int_0^oo dz qquad (sin z)/z = pi/2#
Related questions
Impact of this question
1282 views around the world
You can reuse this answer
Creative Commons License