Suppose $0<k<l$. Show that:
$$\int_{0}^{\infty}e^{-kx}\cos(lx)dx = \frac{k}{k^{2}+l^{2}}$$
I began but doing integration by parts twice and got stuck:
$$-\frac{\cos(lx)e^{-kx}}{k}-\frac{l}{k^{2}}e^{-kx}\sin(lx)-\frac{l^{2}}{k^{2}}\int_{0}^{\infty}e^{-kx}\cos(lx)dx$$
How to evaluate expression further?
We can start by considering the integral: $$\int_{0}^{\infty} e^{-kx} e^{ilx} dx$$ Using the fact that $e^{ix} = \cos x + i \sin x$, we can write this as: $$\int_{0}^{\infty} e^{-(k-il)x} dx = \frac{1}{k-il}$$ Now, we can split the cosine function into its real and imaginary parts: $$\cos(lx) = \frac{e^{ilx} + e^{-ilx}}{2}$$ Substituting this into the integral and using linearity of integration, we get: \begin{align*} \int_{0}^{\infty} e^{-kx} \cos(lx) dx &= \frac{1}{2} \int_{0}^{\infty} e^{-kx} (e^{ilx} + e^{-ilx}) dx \\ &= \frac{1}{2} \left(\int_{0}^{\infty} e^{-(k-il)x} dx + \int_{0}^{\infty} e^{-(k+il)x} dx\right) \\ &= \frac{1}{2} \left(\frac{1}{k-il} + \frac{1}{k+il}\right) \\ &= \frac{1}{2} \cdot \frac{2k}{k^2 + l^2} \\ &= \frac{k}{k^2 + l^2} \end{align*}
There is another solution without using imaginary numbers, which does integration by parts twice.
\begin{aligned} & \int_{0}^{+\infty} \mathrm{e}^{-k x} \cos lx \mathrm{~d} x \\ = & -\frac{1}{k} \int_{0}^{+\infty} \cos lx \mathrm{de}^{-kx} \\ = & -\frac{1}{k}\left(\left.\mathrm{e}^{-kx} \cos lx\right|_{0} ^{+\infty}+l\int_{0}^{+\infty} \mathrm{e}^{-kx} \sin l x \mathrm{~d} x\right) \\ = & -\frac{1}{k}\left(-1-\frac{l}{k} \int_{0}^{+\infty} \sin lx \mathrm{de}^{-kx }\right) \\ = & -\frac{1}{k}\left(-1-\frac lk\left(\left.\mathrm{e}^{-kx} \sin lx\right|_{0} ^{+\infty}-l\int_{0}^{+\infty} \mathrm{e}^{-kx} \cos lx \mathrm{~d} x\right)\right) \\ = & -\frac{1}{k}\left(-1+\frac{l^2}k\int_{0}^{+\infty} \mathrm{e}^{-kx } \cos lx \mathrm{~d} x\right). \end{aligned} Therefore, $$\int_{0}^{+\infty} \mathrm{e}^{-kx } \cos lx \mathrm{~d} x=\frac k{k^2+l^2}$$