Integrating $\int_{0}^{+\infty} e^{-st} \cos(at)\,dt$ using the complex exponential

Question

Other similar questions I have found used integration by part of alternative complicated transformations that I do not understand. Using the $\cos(x) = \frac{1}{2} * (e^{ix} + e^{-ix})$ identity, is there a way to find the solution, which should be $a/(a^2 + s^2)$?

Answer 1

\begin{align}\int_0^{+\infty}e^{-st}\cos(at)\,\mathrm dt&=\left[e^{-st}\frac{\sin(at)}a\right]_{t=0}^{t=+\infty}+\frac sa\int_0^{+\infty}e^{-st}\sin(at)\,\mathrm dt\\&=\left[-\frac s{a^2}e^{-st}\cos(st)\right]_{t=0}^{t=+\infty}-\frac{s^2}{a^2}\int_0^{+\infty}e^{-st}\cos(at)\,\mathrm dt\\&=\frac s{a^2}-\frac{s^2}{a^2}\int_0^{+\infty}e^{-st}\cos(at)\,\mathrm dt\end{align}and therefore$$\int_0^{+\infty}e^{-st}\cos(at)\,\mathrm dt=\frac{\frac s{a^2}}{1+\frac{s^2}{a^2}}=\frac s{s^2+a^2}.$$

---

If you want to use complex numbers, you can do\begin{align}\int_0^{+\infty}e^{-st}\cos(at)\,\mathrm dt&=\frac12\left(\int_0^{+\infty}e^{ati-st}\,\mathrm dt+\int_0^{+\infty}e^{-ati-st}\,\mathrm dt\right)\\&=\frac12\left(\frac{1}{ai-s}+\frac1{-ai-s}\right)\\&=\frac12\times\frac{-2s}{-a^2-s^2}\\&=\frac s{s^2+a^2}.\end{align}

Answer 2

For all complex number $z$,

$$ \cos(z) = \frac{e^{iz} + e^{-iz}}{2}. $$

It follows that:

$$ \ \begin{align*} \int_{0}^{+\infty} e^{-st} \cos(at) \; dt & = {} \frac{1}{2} \int_{0}^{+\infty} \big[ e^{-st} e^{iat} + e^{-st}e^{-iat} \big] \; dt \\[2mm] & = \frac{1}{2} \int_{0}^{+\infty} e^{-(s-ia)t} \; dt + \frac{1}{2} \int_{0}^{+\infty} e^{-(s+ia)t} \; dt. \end{align*} $$

Now, if $z$ is a non-zero complex number,

$$ \int_{0}^{+\infty} e^{-zt} \; dt = \Bigg[ -\frac{1}{z} e^{-zt} \Bigg]_{t=0}^{t=+\infty} = \frac{1}{z}. \tag{1} $$

Therefore, with $z = s - ia$:

$$ \begin{align*} \int_{0}^{+\infty} e^{-st}\cos(at) \; dt & = {} \frac{1}{2z} + \frac{1}{2\overline{z}} \\[2mm] & = \frac{z + \overline{z}}{2 z \overline{z}} \\[2mm] & = \frac{\Re(z)}{\vert z \vert^2} \\[2mm] & = \frac{s}{s^2 + a^2}. \end{align*} $$

Note that $(1)$ is true if $s > 0$ because $\displaystyle \vert e^{-zt} \vert = e^{-\Re(z)t} = e^{-st} \; \rightarrow \; 0$ as $t \to +\infty$.

Answer 3

\begin{align} \int_{0}^{+\infty} e^{-st} \cos(at)\,dt &={\bf Re} \int_{0}^{+\infty}e^{(ia-s)t} dt \\ &={\bf Re} \left(\dfrac{e^{-(s-ia)t}}{s-ia}\right)_{0}^{+\infty} ~~~~;~~~s>0\\ &={\bf Re} \left(\dfrac{1}{s-ia}\right) \\ &=\dfrac{s}{s^2+a^2} \end{align}