KindiakMath

Proving Feynman’s Trick

Feynman’s trick in differentiating under the integral sign has been creatively wielded to evaluate otherwise intractable integrals. In this exercise, we prove Feynman’s trick and use it to evaluate the seemingly intractable Dirichlet integral

\displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x}\, \mathrm dx.

Let (\Omega, \mathcal F, \mathbb R) be a measure space and f : \Omega \times [a, b] \to \mathbb R be a function such that for each t \in [a, b], f( \cdot, t) is measurable.

Problem 1. Suppose the following conditions:

  • For any \omega \in \Omega, f(\omega, \cdot) is continuous.
  • There exists some non-negative integrable g : \Omega \to \mathbb R such that for any (\omega, t) \in \Omega \times [a, b], |f( \omega ,t)| \leq g(\omega).

Prove that the map F : [a, b] \to \mathbb R defined by \displaystyle F(t) : = \int_{\Omega} f(\cdot , t)\, \mathrm d\mu is continuous.

(Click for Solution)

Solution. Fix t_n \to t. For any \omega \in \Omega, since f(\omega, \cdot) is continuous,

f(\omega, t_n) \to f(\omega, t),

so that f(\cdot, t_n) \to f(\cdot, t) pointwise. Furthermore,

\displaystyle \int_{\Omega} |f(\cdot, t)|\, \mathrm d\mu \leq \int_{\Omega} g\, \mathrm d\mu < \infty,

so that f(\cdot, t_n) and f(\cdot, t) are all integrable.

Since g is integrable, by Lebesgue’s dominated convergence theorem,

\displaystyle F(t_n) = \lim_{n \to \infty} \int_{\Omega} f(\cdot , t_n)\, \mathrm d\mu= \int_{\Omega} f(\cdot , t)\, \mathrm d\mu = F(t),

so that F is continuous, as required.

Problem 2. Suppose the following conditions:

  • There exists some t_0 \in [a, b] such that f(\cdot, t_0) is integrable.
  • For each \omega \in \Omega, f(\omega, \cdot) is differentiable with derivative at t_0 denoted by \frac{\partial f}{\partial t}(\omega,t_0).
  • There exists some non-negative integrable g : \Omega \to \mathbb R such that for any (\omega, t) \in \Omega \times [a, b], \left|\frac{\partial f}{\partial t} (\omega,t) \right| \leq g( \omega ).

Prove that the map F : [a, b] \to \mathbb R defined by \displaystyle F(t) : = \int_{\Omega} f(\cdot, t)\, \mathrm d\mu is differentiable on [a, b] and

\displaystyle F'(t) = \frac{\mathrm d}{\mathrm dt} \int_{\Omega} f(\omega, t)\, \mathrm d\mu(\omega) = \int_{\Omega} \frac{\partial f}{\partial t}(\omega, t)\, \mathrm d\mu(\omega).

(Click for Solution)

Solution. We first check that F is well-defined. By hypothesis, F(t_0) is well-defined. Fix t \in [a, b]. By the mean value theorem, there exists c between t_0 and latex t$ such that

\displaystyle \frac{ |f(\omega, t) - f(\omega, t_0)| }{ |t - t_0| } = \left| \frac{ \partial f }{ \partial t} (\omega, c) \right| \leq g(\omega).

By performing more analysis, f(\cdot, t) is integrable, so that F(t) is well-defined.

Now fix \omega \in \Omega. For any t_n \to t, since each f(\cdot, t_n) - f(\cdot, t) is measurable,

\displaystyle \frac{\partial f}{\partial t}(\omega, t) := \lim_{n \to \infty} \underbrace{ \frac{f(\omega, t_n) - f(\omega, t)}{ t_n - t} }_{\varphi_n(\omega)}

is measurable. Furthermore, \varphi_n \to \frac{\partial f}{\partial t} (\cdot, t) pointwise. We claim that |\varphi_n| \leq g, since the mean value theorem gives c between t_n and t such that

\displaystyle \frac{ | f(\omega, t_n) - f(\omega, t) |}{ | t_n - t | } = \left| \frac{\partial f}{\partial t}(\omega, c) \right| \leq g(\omega).

By algebruh and the triangle inequality, each \varphi_n is integrable. Hence, by Lebesgue’s dominated convergence theorem,

\begin{aligned} \lim_{t_n \to t} \int_{\Omega} \frac{f(\omega, t_n) - f(\omega, t)}{t_n - t}\, \mathrm d\mu(\omega) &= \lim_{n \to \infty} \int_{\Omega} \varphi_n\, \mathrm d\mu \\ &= \int_{\Omega} \lim_{n \to \infty} \varphi_n\, \mathrm d\mu = \int_{\Omega} \frac{\partial f}{\partial t} (\cdot, t)\, \mathrm d\mu. \end{aligned}

On the other hand, by bookkeeping

\begin{aligned} \frac{ \mathrm d }{ \mathrm dt } F(t) &= \lim_{ t_n \to t } \frac{ F(t_n) - F(t) }{ t_n - t } \\ &= \lim_{t_n \to t} \int_{\Omega} \frac{f(\omega, t_n) - f(\omega, t)}{t_n - t}\, \mathrm d\mu(\omega) \\ &= \int_{\Omega} \frac{\partial f}{\partial t} (\omega, t)\, \mathrm d\mu(\omega). \end{aligned}

Therefore,

\displaystyle F'(t) = \frac{\mathrm d}{\mathrm dt} \int f(\omega, t)\, \mathrm d\mu(\omega) = \int \frac{\partial f}{\partial t}(\omega, t)\, \mathrm d\mu(\omega).

Remark 1. Thanks to Problem 2, our proof that \mathcal L\{tf'(t)\} = -\mathcal L\{f\}'(s) in the study of differential equations becomes a logically correct one.

Problem 3. Use Problem 2 to evaluate \displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x}\, \mathrm dx.

(Click for Solution)

Solution. Define the function I by

\displaystyle I(y) := \int_{0}^{\infty} \frac{\sin x}{x} \cdot e^{-xy}\, \mathrm dy

that satisfies the hypotheses of Problem 2, and our goal is to evaluate 2 \cdot I(0). Applying Problem 2 and integrating by parts,

\begin{aligned} I'(y) &= \int_{0}^{\infty} \frac{\partial}{\partial y} \left( \frac{\sin x}{x} \cdot e^{-xy} \right) \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{\sin x}{x} \cdot -xe^{-xy} \, \mathrm dx \\ &= -\int_{0}^{\infty} e^{-xy} \cdot \sin x \, \mathrm dx \\ &= -\left[ \frac{e^{-xy}}{(-y)^2 + 1^2} \cdot ((-y) \sin(x) - \cos(x)) \right]_{0}^{\infty} \\ &= -\frac{1}{1+y^2}. \end{aligned}

Integrating and applying the first fundamental theorem of calculus,

\displaystyle I(y) = I(0) - \int_0^y \frac{1}{1+y^2}\, \mathrm dy = I(0) - \tan^{-1}(y).

Since I is continuous,

\displaystyle \lim_{y \to \infty} I(y) = \int_0^\infty \lim_{y \to \infty} \frac{\sin x}{x} \cdot e^{-xy}\, \mathrm dx = \int_0^\infty 0\, \mathrm dx = 0.

Taking y \to \infty on all sides, therefore,

\displaystyle 0 = I(0) - \frac{\pi}{2} \quad \Rightarrow \quad I(0) = \frac{\pi}{2}.

Therefore, the Dirichlet integral evaluates to

\displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x}\, \mathrm dx = 2 \cdot I(0) = \pi.

—Joel Kindiak, 29 Jul 25, 1319H

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