MathematicsFebruary 15, 20263 min read

Evaluating a Beautiful Integral Using Differentiation Under the Integral Sign

A detailed walkthrough of Feynman's favorite technique — differentiating under the integral sign — applied to a classic integral involving logarithms and rational functions.

Introduction

One of the most powerful and elegant techniques in integral calculus is differentiation under the integral sign, sometimes called Leibniz's rule or, more colloquially, Feynman's trick. Richard Feynman famously learned this technique from the book Advanced Calculus by Frederick S. Woods, and it became one of his go-to methods for evaluating difficult integrals.

In this post, we'll use this technique to evaluate the following beautiful integral:

I = \int_0^1 \frac{\ln(1+x)}{1+x^2} \, dx

The Setup

The key idea behind Feynman's trick is to introduce a parameter into the integral, differentiate with respect to that parameter, evaluate the resulting (hopefully simpler) integral, and then integrate back to obtain the original result.

Let us define:

I(a) = \int_0^1 \frac{\ln(1+ax)}{1+x^2} \, dx

where $a \geq 0$ . Notice that:

$I(0) = 0$ (since $\ln(1) = 0$ )
$I(1) = I$ (our target integral)

Differentiating Under the Integral Sign

We differentiate $I(a)$ with respect to $a$ :

I'(a) = \int_0^1 \frac{\partial}{\partial a} \left[ \frac{\ln(1+ax)}{1+x^2} \right] dx = \int_0^1 \frac{x}{(1+ax)(1+x^2)} \, dx

Now we need to evaluate this integral. We use partial fraction decomposition:

\frac{x}{(1+ax)(1+x^2)} = \frac{A}{1+ax} + \frac{Bx + C}{1+x^2}

Multiplying both sides by $(1+ax)(1+x^2)$ and solving for the coefficients, we get:

A = \frac{-a}{1+a^2}, \quad B = \frac{a}{1+a^2}, \quad C = \frac{1}{1+a^2}

Computing the Derivative

Substituting back:

I'(a) = \frac{1}{1+a^2} \int_0^1 \left[ \frac{-a}{1+ax} + \frac{ax+1}{1+x^2} \right] dx

Evaluating each part:

\int_0^1 \frac{-a}{1+ax} \, dx = -\ln(1+a)

\int_0^1 \frac{ax+1}{1+x^2} \, dx = \frac{a}{2}\ln 2 + \frac{\pi}{4}

Therefore:

I'(a) = \frac{1}{1+a^2} \left[ -\ln(1+a) + \frac{a}{2}\ln 2 + \frac{\pi}{4} \right]

Integrating Back

Now we integrate from $0$ to $1$ :

I(1) - I(0) = \int_0^1 I'(a) \, da = \int_0^1 \frac{-\ln(1+a)}{1+a^2} \, da + \frac{\ln 2}{2}\int_0^1 \frac{a}{1+a^2} \, da + \frac{\pi}{4}\int_0^1 \frac{1}{1+a^2} \, da

But notice that $\int_0^1 \frac{\ln(1+a)}{1+a^2} da = I(1) = I$ ! So we have:

I = -I + \frac{\ln 2}{2} \cdot \frac{\ln 2}{2} + \frac{\pi}{4} \cdot \frac{\pi}{4}

2I = \frac{(\ln 2)^2}{4} + \frac{\pi^2}{16}

The Final Result

\boxed{I = \int_0^1 \frac{\ln(1+x)}{1+x^2} \, dx = \frac{\pi^2}{32} + \frac{(\ln 2)^2}{8}}

This is a remarkably clean result! The appearance of both $\pi$ and $\ln 2$ in the answer reflects the deep connections between logarithmic and trigonometric functions that arise naturally in integral calculus.

Key Takeaways

Feynman's trick (differentiation under the integral sign) is a powerful technique for evaluating integrals that resist standard methods.
The technique works best when you can find a natural parameter $a$ such that both $I(0)$ and $I'(a)$ are tractable.
Sometimes the "trick within the trick" is recognizing that the integral you're trying to evaluate appears on both sides of an equation, allowing you to solve for it algebraically.