Here is a cute integral that I like. Let's evaluate: $$I_0 = \int_{-\infty}^{+\infty} e^{-ipx - k |x|} dx$$ We start by seperating the integral so that we can get rid of the absolute value: $$=\int_{-\infty}^{0} e^{-ipx + k x} dx + \int_{0}^{+\infty} e^{-ipx - k x} dx$$ We apply the "product rule of exponents": $$=\int_{-\infty}^{0} e^{-ipx} e^{k x} dx + \int_{0}^{+\infty} e^{-ipx} e^{- k x} dx$$ Expand via Euler's and make use of the fact that sine is an odd function: $$=\int_{-\infty}^{0} [\cos{px} - i \sin{px}] e^{k x} dx + \int_{0}^{+\infty} [\cos{px} - i \sin{px}] e^{- k x} dx$$ We can make the first term be from 0 to positive infinity by making the substitution x -> -x. We use here that cosine is even and sine is odd again: $$=\int_{0}^{+\infty} [\cos{px} + i \sin{px}] e^{- k x} dx + \int_{0}^{+\infty} [\cos{px} - i \sin{px}] e^{- k x} dx$$ Beautiful, now combine the integrals: $$=\int_{0}^{+\infty} [\cos{px} + i \sin{px} +\cos{px} - i \sin{px}] e^{- k x} dx$$ Cancel the sines and bring the 2 (from the addition of the cosines) out front: $$=2\int_{0}^{+\infty} \cos{(px)} e^{- k x} dx$$ Let's define this integral as $I_1$ $$I_1=\int_{0}^{+\infty} \cos{(px)} e^{- k x} dx$$ Let's proceed with integration by parts $$u = \cos{(px)}$$ $$du = -p\sin{(px)} dx$$ $$dv = e^{-kx}$$ $$v = \frac{-1}{k} e^{-kx}$$ $$I_1 = -\frac{\cos{(px)}e^{-kx}}{k}\rvert_{0}^{\infty} - \frac{p}{k} \int_{0}^{+\infty}\sin{(px)}e^{-kx}dx$$ Evaluating: $$I_1 = \frac{1}{k} - \frac{p}{k} \int_{0}^{+\infty}\sin{(px)}e^{-kx}dx$$ Now let's define $I_2$ to be that integrand $$I_2 = \int_{0}^{+\infty}\sin{(px)}e^{-kx}dx$$ Proceed with integration by parts again: $$u = \sin{(px)}$$ $$du = p\cos{(px)}$$ $$dv = e^{-kx}$$ $$v = \frac{-1}{k}e^{-kx}$$ $$I_2 = -\sin{(px)}e^{-kx}/k\rvert_{0}^{\infty} + \frac{p}{k}\int_{0}^{+\infty} \cos{(px)} e^{- k x} dx$$ $$I_2 = \frac{p}{k} I_1$$ We have now obtained a system of equations with $I_1$ $$I_2 = \frac{p}{k} I_1$$ $$I_1 = \frac{1}{k} - \frac{p}{k} I_2$$ Solving for $I_1$ $$I_1 = \frac{1}{k} - \frac{p^2}{k^2} I_1$$ You can do a little bit more algebra after this substitution to get: $$I_1 = \frac{k}{k^2+p^2}$$ Our original integral is twice this $$I_0 = \int_{-\infty}^{+\infty} e^{-ipx - k |x|} dx = \frac{2k}{k^2+p^2}$$