Laplace transform of $\cos(at)$

Question

I need to find the Laplace transform of $\cos(at)$

I know that $L\{\cos(at)\}= \int_{0}^{\infty} e^{-st} \cos (at) dt$ but I am having trouble finding the integral

Thank you

Answer 1

$$\int_0^\infty e^{ct}e^{-st}dt=\int_0^\infty e^{-t(s-c)}=\frac{ e^{-t(s-c)}}{-(s-c)}|_0^{\infty}=\frac1{s-c}\text{ for }s-c>0$$

Putting $c=ib$ $$\int_0^\infty e^{(ib)t}e^{-st}dt=\frac1{s-ib}$$

Using Euler's formula $e^{ix}=\cos x+i\sin x$

$$\int_0^\infty (\cos bt+i\sin bt)e^{-st}dt=\frac1{s-ib}=\frac{s+ib}{s^2+b^2}$$

Equating the real parts,

$$\int_0^\infty (\cos bt) e^{-st}dt =\frac{s }{s^2+b^2}$$

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Alternatively,

$$\int e^{at}\cos btdt=\cos bt\int e^{at}dt-\int\left(\frac{\cos bt}{dt} \int e^{at}dt\right)dt$$ $$=\frac{e^{at}\cos bt}a+\frac ba\int e^{at}\sin bt dt $$

So, $$\int e^{at}\cos btdt=\frac{e^{at}\cos bt}a+\frac ba\int e^{at}\sin bt dt$$

Converting $bt$ to $bt-\frac\pi2$ as $\cos\left(bt-\frac\pi2\right)=\cos\left(\frac\pi2-bt\right)=\sin bt$ and $\sin\left(bt-\frac\pi2\right)=-\sin\left(\frac\pi2-bt\right)=-\cos bt$

$$\int e^{at}\sin btdt=\frac{e^{at}\sin bt}a-\frac ba\int e^{at}\cos bt dt$$

So, $$I=\int e^{at}\cos btdt=\frac{e^{at}\cos bt}a+\frac ba\left(\frac{e^{at}\sin bt}a-\frac ba\int e^{at}\cos bt dt \right)$$

$$\implies I=\frac{e^{at}\cos bt}a+\frac b{a^2}e^{at}{\sin bt}- \frac{b^2}{a^2}I $$

$$\implies \int e^{at}\cos btdt=I=\frac{e^{at}(a\cos bt+b\sin bt)}{a^2+b^2} $$

$$a=-s, \int e^{-st}\cos btdt=I=\frac{e^{-st}(-s\cos bt+b\sin bt)}{s^2+b^2} $$

$$\int_0^\infty (\cos bt) e^{-st}dt =\frac{e^{-st}(-s\cos bt+b\sin bt)}{s^2+b^2}|_0^\infty=\frac s{s^2+b^2} $$

Answer 2

Hint: integrate by parts twice.

Answer 3

Integration by Parts formula: $\displaystyle \int u \ dv = u v - \int v \ du$

Note: The limits are not taken care of here. Make sure you plug them in later.

Let us put $\displaystyle S = \int e^{a x} \sin bx \ dx$ and $\displaystyle C = \int e^{a x} \cos bx \ dx$.

Let us consider the integral $\displaystyle S = \int e^{a x} \sin bx \ dx$

Let $u = \sin bx$ and $dv = e^{ax}$.

Then:

$$S = \int e^{a x} \sin bx \ dx = \frac {e^{a x} \sin bx} a - \int \frac {e^{ax} } a b \cos bx \ dx = \frac {e^{a x} \sin bx} a - \frac b a \int e^{ax} \cos bx \ dx = \frac {e^{a x} \sin bx} a - \frac b a C $$

Now consider $\displaystyle C = \int e^{a x} \cos bx \ dx$:

Let $u = \cos bx$ and $dv = e^{ax}$.

Then:

$$ C = \int e^{a x} \cos bx \ dx =\frac {e^{a x} \cos bx} a - \int \frac {e^{ax} } a \left({-b \sin bx}\right) \ dx =\frac {e^{a x} \cos bx} a + \frac b a \int e^{ax} \sin bx \ dx = \frac {e^{a x} \cos bx} a + \frac b a S $$

Eliminate either $S$ or $C$ and obtain the result required.

$$ S \left({a^2 + b^2}\right)=e^{a x} \left({a \sin bx - b \cos bx}\right) \text{ and }$$ $$ C \left({a^2 + b^2}\right)=e^{a x} \left({a \cos bx + b \sin bx}\right) $$

I leave the limit substitution for you to find out.