\documentclass{amsart}

\title{Differentiating Under the Integral}

\author{Math 166: Honors Calculus II}

\date{Spring, 2000}

\begin{document}

\maketitle

There is a certain technique for evaluating integrals that is no longer
taught in the standard calculus curriculum. It is mentioned in the
autobiography of the renowned physicist Richard Feynman, \emph{Surely
You're Joking Mr.~Feynman}, p.86--87:

\begin{quotation}

   So every physics class, I paid no attention to what was going
on with Pascal's Law, or whatever they were doing. I was up in
the back with this book: \emph{Advanced Calculus}, by Woods. Bader
[Feynman's High School Physics teacher, who loaned Feynman his
copy of Wood's book] knew I had studied \emph{Calculus for the
Practical Man} a little bit, so he gave me the real works---it was
for a junior or senior course in college. It had Fourier series,
Bessel functions, determinants, elliptic functions---all kinds
of wonderful stuff that I didn't know anything about.  

   That book also showed how to differentiate parameters under
the integral sign---it's a certain operation. It turns out that
it's not taught very much in the universities; they don't
emphasize it. But I caught on how to use that method, and I
used that one damn tool again and again. So because I was
self-taught using that book, I had peculiar methods of doing
integrals.

   The result was, when guys at MIT or Princeton had trouble
doing a certain integral, it was because they couldn't do it
with the standard methods they had learned in school. If it was
contour integration, they would have found it; if it was a
simple series expansion, they would have found it. Then I come
along and try differentiating under the integral sign, and often
it worked. So I got a great reputation for doing integrals,
only because my box of tools was different from everybody
else's, and they had tried all their tools on it before giving
the problem to me.

\end{quotation}

The method is illustrated by the following example. To evaluate
  \[ \int_0^1 \frac{x^5 - 1}{\log(x)}\,dx \]
first introduce a parameter, say $\alpha$, into the integral, thereby
creating a function of this parameter.
  \[ f(\alpha) = \int_0^1 \frac{x^\alpha - 1}{\log(x)}\,dx \]
The problem now is to find $f(5)$. This will be done by computing
$f'(\alpha)$---taking the derivative with respect to $\alpha$ inside
the integral---then integrating with respect to~$x$ to get an explicit
expression for $f'(\alpha)$. Finally, integrating this expression with
respect to~$\alpha$ gives an expression for $f(\alpha)$:
  \[ f'(\alpha) = \int_0^1 \frac{d}{d\alpha} \Big(\frac{x^\alpha -
  1}{\log(x)}\Big)\,dx = \int_0^1 \frac{x^\alpha\log(x)}{\log(x)}\,dx
  = \int_0^1 x^\alpha\,dx = \frac{x^{\alpha+1}}{\alpha+1}\Big|_0^1 =
  \frac{1}{\alpha+1} \]
Now integrating with respect to~$\alpha$ gives
  \[ f(\alpha) = \int \frac{1}{\alpha+1}\,d\alpha = \log(\alpha+1) + c \]
Since $f(0) = 0$, we see that $c=0$ and $f(\alpha) = \log(\alpha+1)$. The
answer to our original problem is then $f(5) = \log(6)$. In fact, we
have derived the general formula
  \[ \int_0^1 \frac{x^\alpha - 1}{\log(x)}\,dx = \log(\alpha+1) \]

The method of ``differentiating under the integral'' appears in many old
calculus texts, mainly those published before 1945. See, for example,
Edwin Wilson's text, \emph{Advanced Calculus}, published in 1912,
where examples and a rigorous justification for the method are given
on p.281--288.

Here are a few other definite integrals that can be evaluated
in this manner.
  \[ \int_0^\pi \log(1 + \alpha\cos(x))\,dx =
  \pi\log\frac{1+\sqrt{1-\alpha^2}}{2} \]
  \[ \int_0^\pi \log(1 -2u\cos(x)+\alpha^2)\,dx =
  \left\{\begin{array}{ll} \pi\log \alpha^2 & \alpha^2\ge1\cr 0 &
  \alpha^2\le1\cr\end{array}\right. \]
  \[ \int_0^{\pi/2} \frac{\log(1 + \cos(\alpha)\cos(x))}{\cos(x)}\,dx =
  \frac{1}{2}\Big(\frac{\pi^2}{4} - \alpha^2\Big) \]
  \[ \int_0^1 x^\alpha(\log(x))^n\,dx = (-1)^n\frac{n!}{(\alpha+1)^{n+1}} \]
  \[ \int_0^\infty\frac{dx}{(x^2 + \alpha^2)^{n+1}} = \frac{\pi}{2}
  \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n\cdot\alpha^{2n+1}} \]
Keep in mind that to successfully apply this method some experimentation
may be needed to discover an appropriate way to introduce a parameter
into the integral.  So, for example, to evaluate
  \[ \int_0^{\pi/2} \frac{\log(1 + (1/2)\cos(x))}{\cos(x)}\,dx \]
it may take some time to realize that replacing $1/2$ by
$\cos(\alpha)$ leads to a solution. The answer, according
to the above formula with $\alpha=\frac{\pi}{3}$, is
$\frac{1}{2}\Big(\frac{\pi^2}{4}-\frac{\pi^2}{9}\Big)=\frac{5\pi^2}{72}$.

\end{document}
