Feynmann's integral trick

feynmann_integral_trick.tex

\documentclass{article}
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\title{Feynmann's integral trick}
\date{August 2018}
\begin{document}
\maketitle

The trick is to define a new function $I \colon \mathbb{R} \to \mathbb{R}$ such that $I(a_0)$ is the
integral we are trying to evaluate and $I(a_1)$ reduces to something more easily evaluated. The next
step is to derive $I(a)$ w.r.t $a$. With a bit of luck, you end up in a situation that you have a
differential equation for $I(a)$, and a fixed boundary condition at $I(a_1)$.

Here is an example from Putnam:

\begin{align*}
  \int_0^1 \frac{\log(x + 1)}{x^2 + 1} dx
\end{align*}

First, we defined $I(a)$, keeping in mind that there has to be some $a_0$ such that $I(a_0)$ is our
initial integral and some $a_1$ such that $I(a_1)$ is easy. Thinking along the lines of the latter,
we'd be fine if somehow the numerator went away.

\begin{align*}
  I(a) = \int_0^1 \frac{\log(ax + 1)}{x^2 + 1} dx
\end{align*}

Then, we see that the initial integral is $I(1)$. Next, we take the derivative of $I(a)$ w.r.t. $a$:

\begin{align*}
  \frac{\partial I(a)}{\partial a}
    = \int_0^1 \frac{\partial}{\partial a} \frac{\log(ax + 1)}{x^2 + 1} dx
    = \int_0^1 \frac{x}{(x^2 + 1)(ax + 1)} dx
\end{align*}

This has turned into a partial fractions problem. With a bit of linear algebra, we can break down
the rational polynomial into simpler components.

\begin{align*}
  \frac{x}{(x^2 + 1)(ax + 1)}
    = \frac{1}{a^2 + 1} \left( \frac{x}{x^2 + 1} + \frac{a}{x^2 + 1} - \frac{a}{ax + 1} \right)
\end{align*}

Each of these summands has simple antiderivatives.

\begin{align*}
  \int_0^1 \frac{x}{(x^2 + 1)(ax + 1)} dx
    &= \frac{1}{a^2 + 1} \left( \int_0^1 \frac{x}{x^2 + 1} dx
                              + \int_0^1 \frac{a}{x^2 + 1} dx
                              - \int_0^1 \frac{a}{ax + 1} dx
                        \right) \\
    &= \frac{1}{a^2 + 1} \left( \frac{1}{2} \int_0^1 \frac{2x}{x^2 + 1} dx
                              + a           \int_0^1 \frac{1}{x^2 + 1}  dx
                              -             \int_0^1 \frac{a}{ax + 1}   dx
                        \right) \\
    &= \frac{1}{a^2 + 1} \left( \frac{1}{2} \left[ \log(x^2 + 1) \right]_{x=0}^1
                              + a           \left[ \tan^{-1}(x)  \right]_{x=0}^1
                              -             \left[ \log(ax + 1)  \right]_{x=0}^1
                        \right) \\
    &= \frac{1}{a^2 + 1} \left( \frac{\log 2}{2}
                              + \frac{\pi a}{4}
                              - \log(a + 1)
                        \right)
\end{align*}

Now, recall that all of this is the derivative of $I(a)$ w.r.t. $a$, so $I(a)$ is an antiderivative
of the above expression.

\begin{align*}
  \frac{\partial I(a)}{\partial a}
    &= \frac{1}{a^2 + 1} \left( \frac{\log 2}{2} + \frac{\pi a}{4} - \log(a + 1) \right)
\end{align*}

At this point, we just integrate. We choose our lower bound to be something where $I$ is trivial. In
this case, $I(0) = 0$.

\begin{align*}
  I(a)
    &= I(0) + \int_0^a \left[ \frac{\log 2}{2} \cdot \frac{1}{\alpha^2 + 1}
                            + \frac{\pi}{8} \cdot \frac{2\alpha}{\alpha^2 + 1}
                            - \frac{\log(\alpha + 1)}{\alpha^2 + 1}
                      \right] d\alpha \\
    &= 0 + \left[ \frac{\log 2}{2} \int_0^a \frac{1}{\alpha^2 + 1}      d\alpha
                + \frac{\pi}{8}    \int_0^a \frac{2a}{a^2 + 1}          d\alpha
                -                  \int_0^a \frac{\log(a + 1)}{a^2 + 1} d\alpha
                          \right] \\
    &=   \frac{\log 2}{2} \left[ \tan^{-1}(\alpha) \right]_{\alpha=0}^a
       + \frac{\pi}{8}    \left[ \log(\alpha^2 + 1) \right]_{\alpha=0}^a
       -                  \int_0^a \frac{\log(\alpha + 1)}{\alpha^2 + 1} d\alpha \\
    &=    \frac{\pi\log 2}{4}
       -  \int_0^a \frac{\log(\alpha + 1)}{\alpha^2 + 1} d\alpha \\
\end{align*}

Finally, we tie the knot, by setting $a = 1$.

\begin{align*}
  I(1)
     =    \frac{\pi\log 2}{4}
       -  \int_0^1 \frac{\log(\alpha + 1)}{\alpha^2 + 1} d\alpha
     = \frac{\pi\log 2}{4} - I(1)
\end{align*}

It follows that $I(1) = \int_0^1 \frac{\log(x + 1)}{x^2 + 1} dx = \frac{\pi \log 2}{8}$. \hfill
$\square$ 

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