Feynman Integration

Problem: Calculate the integral

$$\int\limits_{0}^{\infty} \frac{\sin x}{x}\mathrm{d}x$$

Feynman’s method: Solve the following equation for $t$

$$I(t) = \int\limits_{0}^{\infty} \frac{\sin x}{x} e^{-tx}\mathrm{d}x$$

and calculate $I(0)$.

The trick here is to differentiate $I(t)$ under the integral sign and solve the differential equation.

$$\begin{align*} I'(t) & = \int\limits_{0}^{\infty} \frac{\partial}{\partial t} \left( \frac{\sin x}{x} e^{-tx} \right) \mathrm{d}x \\ & = - \int\limits_{0}^{\infty} e^{-tx}\sin{x} \;\mathrm{d}x = \frac{1}{t}\int\limits_{0}^{\infty} \sin{x} \;\mathrm{d} e^{-tx} \\ & = \frac{1}{t}\left[ \left. e^{-tx} \sin{x} \right|_{0}^{\infty} - \int\limits_{0}^{\infty} e^{-tx} \mathrm{d} \sin{x} \right] \\ & = - \frac{1}{t} \int\limits_{0}^{\infty} e^{-tx} \cos{x} \;\mathrm{d} x = \frac{1}{t^2} \int\limits_{0}^{\infty} \cos{x} \;\mathrm{d} e^{-tx} \\ & = \frac{1}{t^2}\left[ \left. e^{-tx} \cos{x} \right|_{0}^{\infty} - \int\limits_{0}^{\infty} e^{-tx} \mathrm{d} \cos{x} \right] \\ & = \frac{1}{t^2}\left[ -1 + \int\limits_{0}^{\infty} e^{-tx} \sin{x} \;\mathrm{d}x \right] = \frac{1}{t^2} \left( -1 - I'(t) \right) \end{align*}$$

And here is the differential equation for $I(t)$

$$I'(t) = \frac{-1}{t^2 + 1}$$

which can be solved by integration, using the substitution method.

$$\begin{align*} I(t) & = - \int \frac{\mathrm{d}t}{t^2 + 1} \\ & = \left| t = \tan{y}, \quad \mathrm{d}t = \frac{\mathrm{d}y}{\cos^2{y}}, \quad t^2 + 1 = \frac{1}{\cos^2{y}} \right| \\ & = - \int \mathrm{d}y = -y + C = C - \arctan{t} \end{align*}$$

We can find the constant $C$ by computing $I(+\infty)$

$$0 = \int\limits_{0}^{\infty} \frac{\sin x}{x} e^{-\infty x}\mathrm{d}x = I(+\infty) = C - \arctan(+\infty) = C - \frac{\pi}{2}$$

The final solution for $I(t)$ is

$$I(t) = \frac{\pi}{2} - \arctan{t}$$

We can now calculate the original integral

$$\int\limits_{0}^{\infty} \frac{\sin x}{x}\mathrm{d}x = I(0) = \frac{\pi}{2} - \arctan(0) = \frac{\pi}{2}$$