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

## All I know is $\frac{a}{{a}^{2} + {b}^{2}}$, and you have to use the below limit four times. $\setminus {\lim}_{x \to \setminus \infty} {e}^{- a x} = 0$

Mar 22, 2018

${\int}_{0}^{\infty} {e}^{- a x} \cos b x \mathrm{dx} = \frac{a}{{a}^{2} + {b}^{2}}$

#### Explanation:

Integrate by parts:

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

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

Now:

${\lim}_{x \to \infty} {e}^{- a x} \sin b x = 0$

and ${e}^{0} \sin \left(0\right) = 0$ so:

${\left[\frac{{e}^{- a x} \sin b x}{b}\right]}_{0}^{\infty} = 0$

then:

${\int}_{0}^{\infty} {e}^{- a x} \cos b x \mathrm{dx} = \frac{a}{b} {\int}_{0}^{\infty} {e}^{- a x} \sin b x \mathrm{dx}$

Integrate by parts again:

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

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

Again for $x \to \infty$:

${\lim}_{x \to \infty} {e}^{- a x} \cos b x = 0$

and for $x = 0$

${e}^{0} \cos \left(0\right) = 1$

then:

${\int}_{0}^{\infty} {e}^{- a x} \cos b x \mathrm{dx} = \frac{a}{b} ^ 2 - {a}^{2} / {b}^{2} {\int}_{0}^{\infty} {e}^{- a x} \cos b x \mathrm{dx}$

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

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

$\frac{{b}^{2} + {a}^{2}}{b} ^ 2 {\int}_{0}^{\infty} {e}^{- a x} \cos b x \mathrm{dx} = \frac{a}{b} ^ 2$

${\int}_{0}^{\infty} {e}^{- a x} \cos b x \mathrm{dx} = \frac{a}{{a}^{2} + {b}^{2}}$

Mar 23, 2018

$\frac{a}{{a}^{2} + {b}^{2}}$

#### Explanation:

Using the de Moivre's identity

${e}^{i b x} = \cos \left(b x\right) + i \sin \left(b x\right)$ we have

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

but

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

now if $a > 0$ we have

$\frac{- a - i b}{{a}^{2} + {b}^{2}} \left(0 - 1\right) = \frac{a + i b}{{a}^{2} + {b}^{2}}$

Finally

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