Feynman
Drew Allan Bertwistle
Feynman talks vaguely about integration techniques and he mentions
'differentiating under the integral sign'
does anyone know specifically what technique he may have been refering to?
Doug Magnoli
I can answer your question, but then I have another thereto appertaining.
The following is taken from Shaum's Outline of Advanced Calculus, a decent
review sort of book, but hardly a good primary source of information.
Leibnitz's rule is generally expressed something like this:
Let g(a) = Int[u1,u2,f(x,a) dx]
where u1 and u2 may depend on a. Then:
dg/da = Int[u1,u2,df/da dx] + f[u2,a] * du2/da - f[u1,a] * du1/da
with all the usual stipulations about f being differentiable and continuous
etc.
You can see that if u1 and u2 don't depend on a, we have merely the
transposition of the differentiation and the integration, hence
'differentiating under the integral sign.'
An example of its utility:
Evaluate Int[0,1,(x-1)/ln(x) dx]
Define g(a) = Int [0,1,(x^a-1)/ln(x) dx] y>0
Then g'(a) = Int[0,1, (d/da) ((x^a-1)/ln(x)) dx
= Int [0,1,(x^a ln(x))/ln(x) dx]
= Int [0,1, x^a dx] = 1/(a+1)
Still from Schaum's, "Integrating with respect to a, g(a) = ln(a+1) + c. but
since g(0)=0, c=0 and so g(a)=ln(a+1). then the value of the required
integral is g(1)=ln 2."
So the trick here was to introduce another parameter, a, then take the
derivative of the integral with respect to a. Switching the order of the
differentiation and the integration gives an integral of something that's
straightforward. Now integrate this wrt a to get g(a), and substitute in the
original value for a to get the answer.
I'll give one other example from Schaum's problems:
Find Int[0,1, ln(x)*x^p dx], p>-1
The trick here is to recognize that ln(x)*x^p is d/dp (x^p). So:
Int[0,1, x^p dx] = x^(p+1)/(p+1) eval from 0,1 = 1/(p+1)
d/dp Int[0,1,x^p dx] = d/dp (1/(p+1)) = -1/(p+1)^2
= Int [0,1, d/dp (x^p) dx] = Int [0,1, ln(x)*x^p dx]
And there you have it....Int[0,1,ln(x)*x^p dx] = -1/(p+1)^2.
Schaum gives a couple more examples, most of which I used to be able to solve
but which are eluding me tonight. E.g.,
Prove that Int[0,pi, ln(1+a*cos(x) dx] = pi*ln((1+sqrt(1-a^2))/2)
Prove that Int[0,pi,ln(1-2*a*cos(x)+a^2) dx] = pi*ln(a^2) for |a|<1, 0 for
|a|>1
Show that Int[0,pi, 1/(5-3*cos(x))^3 dx] = 59*pi/2048
What I've wondered since reading that book and finding out what
'differentiating under the integral sign' means is how are you supposed to
recognize the cases where this is going to help? You have to be able to
notice that something about the integrand, when differentiated wrt a variable
other than the variable over which it's to be integrated, will give you
something you can integrate. But it's a technique I've never been able to
use successfully (other than on those Shaum's problems which are currently
eluding me). I've checked out several books from the technical library at
work, all fo which include differentiation under the integral sign in the
index, and all of them basically provide Liebnitz's rule, and that's it.
It's application I'm interested in at this point, but can't find out how to
use it.
-Doug Magnoli
Dave L. Renfro
[sci.math Sat, 08 Apr 2000 05:06:30 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/wehelcrerm>
wrote
> In 'Surely You're Joking Mr. Feynman'
>
> Feynman talks vaguely about integration techniques and he mentions
> 'differentiating under the integral sign'
>
> does anyone know specifically what technique he may have been
> refering to?
I've included an excerpt from Feynman's book where this matter
is mentioned at the end of my post.
Here is an example of the method Feynman mentions. Let's say
you want to evaluate
INTEG(x=0 to x=1) of (x^10 - 1) / (ln x).
To evaluate this, we'll evaluate something that appears to be
even more difficult:
INTEG(x=0 to x=1) of (x^b - 1) / (ln x) for b >= 0.
Consider the following function of b: F(b) is the value of
the integral above that involves b. Ignoring the matter of
interchanging the order of integration with differentiation
(this is justified in the reference I give below), we have
F'(b) = (d/db)(the integral above involving b)
= INTEG(x=0 to x=1) of (d/db)[(x^b - 1) / (ln x)]
= INTEG(x=0 to x=1) of x^b
= (b+1)^(-1).
Integrating this with respect to b gives F(b) = ln(b+1) + C
for some constant C. We can find C by using a value of b for
which we know the value of F. Clearly, F(0) = 0. This implies
that C = 0, and so F(b) = ln(b+1). Hence, the first integral
I gave has the value F(10) = ln(11).
You can find this method discussed in older (say, before 1945)
advanced calculus texts. Feynman learned it from a book by
Woods. I don't have a copy of Woods' book, so I can't give you
a specific reference in it. However, it can also be found on
pp. 281-288 (Section 118: Integrals Containing a Parameter) of
Edwin Bidwell Wilson, ADVANCED CALCULUS, Ginn and Company, 1912.
Here are two other definite integrals that can be evaluated
in this manner:
INTEG(x=0 to Pi) of ln[1 + b*cos(x)]
INTEG(x=0 to Pi/2) of ln[1 + (cos b)(cos x)] / (cos x)
I can easily imagine that if Feynman were asked to evaluate
INTEG(x=0 to Pi/2) of ln[1 + (1/2)(cos x)] / (cos x),
then he could very quickly "test out" various ways of introducing
a parameter (replace 1/2 with b, etc.) until he found a way that
works (replace 1/2 with cos b). The value of this integral, by
the way, is (5/72)*Pi^2. (Exercise for sci.math. Note that this
is an improper integral.)
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Excerpt from Richard Feynman, SURELY YOU'RE JOKING
MR. FEYNMAN, W. W. Norton & Company, 1985 (pp. 86-87,
towards the end of the section titled "A Different Box
of Tools"). Text enclosed in brackets, [...], are NOT
from the original. (These are additions I've included.)
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: "Advanced Calculus", by Woods. Bader
[Feynman's High School Physics teacher, who loaned Feynman his
copy of Wood's book] knew I had studied "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.
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Dave L. Renfro