Question

Find `\int_0^ooe^(-ax)\sin(bx)dx`?

I know these:

• `e^-(ax)=\sum_(n=0)^oo((-ax)^n)/(n!)`
• `\sin(bx)=\sum_(n=0)^oo(-1)^n((bx)^(2n+1))/((2n+1)!)`

However... (see my answer below) can I do that?

Answer 1

See Explanation

Explanation:

`f(x)=\sum_(n=0)^oo(-ax)^n/(n!)(-1)^n((bx)^(2n+1))/((2n+1)!)`
`...=\sum_0^oo((-a)^nx^n)/(n!)(-1)^n(b^(2n+1)x^(2n+1))/((2n+1)!)`
`...=\sum_0^oo((-a)^nb^(2n+1))/(n!(2n+1)!)(-1)^n\int_0^\inftyx^nx^(2n+1)dx...`

Answer 2

`int_0^oo \ e^(-ax) \ sin(bx) \ dx = b/(a^2+b^2)`

Explanation:

We seek:

`I = int_0^oo \ e^(-ax) \ sin(bx) \ dx`

We can apply Integration By Parts:

Let `{ (u,=e^(-ax), => (du)/dx,=-ae^(-ax)), ((dv)/dx,=sin(bx), => v,=-1/bcos(bx) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int_0^oo \ (e^(-ax))(sin(bx)) \ dx = [(e^(-ax))(-1/bcos(bx))]_0^oo - int_0^oo \ (-1/bcos(bx))(-ae^(-ax)) \ dx`

`:. I = -1/b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a/b \ int_0^oo \ e^(-ax) \ cos(bx) \ dx`

Now, Consider the second integral, For which we apply a second application of Integration By Parts:

Let `{ (u,=e^(-ax), => (du)/dx,=-ae^(-ax)), ((dv)/dx,=cos(bx), => v,=1/bsin(bx) ) :}`

So that:

`int_o^oo \ (e^(-ax))(cos(bx)) \ dx = [ \ (e^(-ax))(1/bsin(bx)) \ ]_o^oo - int_o^oo \ (1/bsin(bx))(-ae^(-ax)) \ dx`

`:. int_o^oo \ e^(-ax) \ cos(bx) \ dx = 1/b \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo +a/b \ I`

And, Combining these results, we get:

`I = -1/b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a/b {1/b \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo +a/b \ I}`

`:. I = -1/b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a/b^2 \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo - a^2/b^2 \ I`

`:. b^2I = -b \ [ \ e^(-ax) \ cos(bx) \ ]_0^oo - a \ [ \ e^(-ax) \ sin(bx) \ ]_o^oo - a^2 \ I`

`:. (a^2+b^2)I = -b \ L_1 - a \ L_2`

Where:

`L_1 = [ e^(-ax) \ cos(bx) ]_0^oo`

`\ \ \ \ = (lim_(x rarr oo) e^(-ax) \ cos(bx)) - (e^0cos0)`

`\ \ \ \ = (0) - (1)`

`\ \ \ \ = -1`

And:

`L_2 = [ \ e^(-ax) \ sin(bx) \ ]_o^oo`

`\ \ \ \ = (lim_(x rarr oo) e^(-ax) \ sin(bx)) - (e^0sin0)`

`\ \ \ \ = (0) - (0)`

`\ \ \ \ = 0`

So that:

`(a^2+b^2)I = b => I = b/(a^2+b^2)`

Answer 3

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

Explanation:

Alternatively, using complex numbers:

`int_0^oo e^(-ax)sin(bx)dx = int_0^oo e^(-ax)(e^(ibx)-e^(-ibx))/(2i)dx`

`int_0^oo e^(-ax)sin(bx)dx = 1/(2i)int_0^oo e^(-ax+ibx)dx-1/(2i)int e^(-a-ibx)dx`

`int_0^oo e^(-ax)sin(bx)dx = 1/(2i)[ e^(-ax+ibx)/(-a+ib) +e^(-ax-ibx)/(a+ib)]_0^oo`

`int_0^oo e^(-ax)sin(bx)dx = 1/(2i)[ e^(-ax)(e^(ibx)/(-a+ib) +e^(-ibx)/(a+ib))]_0^oo`

`int_0^oo e^(-ax)sin(bx)dx = -1/(2i)[ e^(-ax)((a+ib)e^(ibx)+(-a+ib) e^(-ibx))/((a-ib)(a+ib))]_0^oo`

`int_0^oo e^(-ax)sin(bx)dx = -1/(2i)[ e^(-ax)(ae^(ibx)+ibe^(ibx)-ae^(-ibx)+ ibe^(-ibx))/(a^2+b^2)]_0^oo`

`int_0^oo e^(-ax)sin(bx)dx = -1/(2i)[ e^(-ax)/(a^2+b^2)(a(e^(ibx)-e^(-ibx))+ib(e^(ibx)+e^(-ibx)))]_0^oo`

`int_0^oo e^(-ax)sin(bx)dx = -[ e^(-ax)/(a^2+b^2)(a((e^(ibx)-e^(-ibx)))/(2i)+b((e^(ibx)+e^(-ibx)))/2)]_0^oo`

`int_0^oo e^(-ax)sin(bx)dx = -[ e^(-ax)/(a^2+b^2)(asin(bx)+bcos(bx))]_0^oo`

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