\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

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

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)