# 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?

Jun 5, 2018

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...

Jun 5, 2018

${\int}_{0}^{\infty} \setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus \mathrm{dx} = \frac{b}{{a}^{2} + {b}^{2}}$

#### Explanation:

We seek:

$I = {\int}_{0}^{\infty} \setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus \mathrm{dx}$

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:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

We have:

${\int}_{0}^{\infty} \setminus \left({e}^{- a x}\right) \left(\sin \left(b x\right)\right) \setminus \mathrm{dx} = {\left[\left({e}^{- a x}\right) \left(- \frac{1}{b} \cos \left(b x\right)\right)\right]}_{0}^{\infty} - {\int}_{0}^{\infty} \setminus \left(- \frac{1}{b} \cos \left(b x\right)\right) \left(- a {e}^{- a x}\right) \setminus \mathrm{dx}$

$\therefore I = - \frac{1}{b} \setminus {\left[\setminus {e}^{- a x} \setminus \cos \left(b x\right) \setminus\right]}_{0}^{\infty} - \frac{a}{b} \setminus {\int}_{0}^{\infty} \setminus {e}^{- a x} \setminus \cos \left(b x\right) \setminus \mathrm{dx}$

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:

${\int}_{o}^{\infty} \setminus \left({e}^{- a x}\right) \left(\cos \left(b x\right)\right) \setminus \mathrm{dx} = {\left[\setminus \left({e}^{- a x}\right) \left(\frac{1}{b} \sin \left(b x\right)\right) \setminus\right]}_{o}^{\infty} - {\int}_{o}^{\infty} \setminus \left(\frac{1}{b} \sin \left(b x\right)\right) \left(- a {e}^{- a x}\right) \setminus \mathrm{dx}$

$\therefore {\int}_{o}^{\infty} \setminus {e}^{- a x} \setminus \cos \left(b x\right) \setminus \mathrm{dx} = \frac{1}{b} \setminus {\left[\setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus\right]}_{o}^{\infty} + \frac{a}{b} \setminus I$

And, Combining these results, we get:

$I = - \frac{1}{b} \setminus {\left[\setminus {e}^{- a x} \setminus \cos \left(b x\right) \setminus\right]}_{0}^{\infty} - \frac{a}{b} \left\{\frac{1}{b} \setminus {\left[\setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus\right]}_{o}^{\infty} + \frac{a}{b} \setminus I\right\}$

$\therefore I = - \frac{1}{b} \setminus {\left[\setminus {e}^{- a x} \setminus \cos \left(b x\right) \setminus\right]}_{0}^{\infty} - \frac{a}{b} ^ 2 \setminus {\left[\setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus\right]}_{o}^{\infty} - {a}^{2} / {b}^{2} \setminus I$

$\therefore {b}^{2} I = - b \setminus {\left[\setminus {e}^{- a x} \setminus \cos \left(b x\right) \setminus\right]}_{0}^{\infty} - a \setminus {\left[\setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus\right]}_{o}^{\infty} - {a}^{2} \setminus I$

$\therefore \left({a}^{2} + {b}^{2}\right) I = - b \setminus {L}_{1} - a \setminus {L}_{2}$

Where:

${L}_{1} = {\left[{e}^{- a x} \setminus \cos \left(b x\right)\right]}_{0}^{\infty}$

$\setminus \setminus \setminus \setminus = \left({\lim}_{x \rightarrow \infty} {e}^{- a x} \setminus \cos \left(b x\right)\right) - \left({e}^{0} \cos 0\right)$

$\setminus \setminus \setminus \setminus = \left(0\right) - \left(1\right)$

$\setminus \setminus \setminus \setminus = - 1$

And:

${L}_{2} = {\left[\setminus {e}^{- a x} \setminus \sin \left(b x\right) \setminus\right]}_{o}^{\infty}$

$\setminus \setminus \setminus \setminus = \left({\lim}_{x \rightarrow \infty} {e}^{- a x} \setminus \sin \left(b x\right)\right) - \left({e}^{0} \sin 0\right)$

$\setminus \setminus \setminus \setminus = \left(0\right) - \left(0\right)$

$\setminus \setminus \setminus \setminus = 0$

So that:

$\left({a}^{2} + {b}^{2}\right) I = b \implies I = \frac{b}{{a}^{2} + {b}^{2}}$

Jun 5, 2018

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = \frac{b}{{a}^{2} + {b}^{2}}$

#### Explanation:

Alternatively, using complex numbers:

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = {\int}_{0}^{\infty} {e}^{- a x} \frac{{e}^{i b x} - {e}^{- i b x}}{2 i} \mathrm{dx}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = \frac{1}{2 i} {\int}_{0}^{\infty} {e}^{- a x + i b x} \mathrm{dx} - \frac{1}{2 i} \int {e}^{- a - i b x} \mathrm{dx}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = \frac{1}{2 i} {\left[{e}^{- a x + i b x} / \left(- a + i b\right) + {e}^{- a x - i b x} / \left(a + i b\right)\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = \frac{1}{2 i} {\left[{e}^{- a x} \left({e}^{i b x} / \left(- a + i b\right) + {e}^{- i b x} / \left(a + i b\right)\right)\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = - \frac{1}{2 i} {\left[{e}^{- a x} \frac{\left(a + i b\right) {e}^{i b x} + \left(- a + i b\right) {e}^{- i b x}}{\left(a - i b\right) \left(a + i b\right)}\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = - \frac{1}{2 i} {\left[{e}^{- a x} \frac{a {e}^{i b x} + i b {e}^{i b x} - a {e}^{- i b x} + i b {e}^{- i b x}}{{a}^{2} + {b}^{2}}\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = - \frac{1}{2 i} {\left[{e}^{- a x} / \left({a}^{2} + {b}^{2}\right) \left(a \left({e}^{i b x} - {e}^{- i b x}\right) + i b \left({e}^{i b x} + {e}^{- i b x}\right)\right)\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = - {\left[{e}^{- a x} / \left({a}^{2} + {b}^{2}\right) \left(a \frac{\left({e}^{i b x} - {e}^{- i b x}\right)}{2 i} + b \frac{\left({e}^{i b x} + {e}^{- i b x}\right)}{2}\right)\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = - {\left[{e}^{- a x} / \left({a}^{2} + {b}^{2}\right) \left(a \sin \left(b x\right) + b \cos \left(b x\right)\right)\right]}_{0}^{\infty}$

${\int}_{0}^{\infty} {e}^{- a x} \sin \left(b x\right) \mathrm{dx} = \frac{b}{{a}^{2} + {b}^{2}}$