Question

`\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`

Answer 1

`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)`

Answer 2

`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)`