#\int_0^\inftye^(-ax)\cos(bx)dx#?

All I know is #a/(a^2+b^2)#, and you have to use the below limit four times.

#\lim_(x->\infty)e^(-ax)=0#

2 Answers
Mar 22, 2018

Answer:

# int_0^oo e^(-ax)cos bx dx = a/(a^2+b^2)#

Explanation:

Integrate by parts:

#int_0^oo e^(-ax)cos bx dx = int_0^oo e^(-ax)d(1/bsin bx)#

#int_0^oo e^(-ax)cos bx dx = [(e^(-ax)sinbx)/b]_0^oo - 1/b int_0^oo sinbxd(e^(-ax))#

Now:

#lim_(x->oo) e^(-ax)sinbx = 0#

and #e^0sin(0) = 0# so:

#[(e^(-ax)sinbx)/b]_0^oo = 0#

then:

#int_0^oo e^(-ax)cos bx dx = a/b int_0^oo e^(-ax)sinbxdx#

Integrate by parts again:

#int_0^oo e^(-ax)cos bx dx = a/b int_0^oo e^(-ax)d(-1/b cos bx)#

#int_0^oo e^(-ax)cos bx dx = [-a/b^2 e^(-ax)cosbx]_0^oo + a/b^2 int_0^oo cos bx d(e^(-ax))#

Again for #x->oo#:

#lim_(x->oo) e^(-ax)cosbx = 0#

and for #x=0#

#e^0cos(0) = 1#

then:

#int_0^oo e^(-ax)cos bx dx = a/b^2 - a^2/b^2 int_0^oo e^(-ax)cos bx dx#

The integral now appears on both sides of the equation and we can solve for it:

#(1+a^2/b^2) int_0^oo e^(-ax)cos bx dx = a/b^2#

#(b^2+a^2)/b^2 int_0^oo e^(-ax)cos bx dx = a/b^2#

# int_0^oo e^(-ax)cos bx dx = a/(a^2+b^2)#

Mar 23, 2018

Answer:

#a/(a^2+b^2)#

Explanation:

Using the de Moivre's identity

#e^(ibx) = cos(bx)+i sin(bx)# we have

#int_0^oo e^(-ax)cos(bx) dx = "Re"[int_0^oo e^((-a+ib)x) dx]#

but

#int_0^oo e^((-a+ib)x) dx = 1/(-a+ib)[e^((-a+ib)x)]_0^oo = (-a-ib)/(a^2+b^2)[e^((-a+ib)x)]_0^oo#

now if #a > 0# we have

#(-a-ib)/(a^2+b^2)(0-1) = (a+ ib)/(a^2+b^2)#

Finally

#int_0^oo e^(-ax)cos(bx) dx =a/(a^2+b^2)#