Integrating by integrating under the integral sign — the other Feynman trick?

$\begingroup$

Having been introduced to the Feynman technique of integration, it seemed natural to wonder if it could be done the other way:

  1. Introduce a new parameter $a$

  2. Integrate with respect to $a$

  3. Integrate with respect to the variable $x$

  4. Differentiate with respect to $a$

  5. Set $a=1$ and add a constant by hand

For instance,

\begin{align} \int x \cos(x) dx =& \frac{d}{da}\int \int x \cos(ax) da dx\\ =& \frac{d}{da}\int \frac{x \sin(ax)}{x}dx\\ =& \frac{d}{da}\int \sin(ax)dx\\ =& \frac{d}{da}\frac{-\cos(ax)}{a}\\ =& \frac{x \sin(ax)}{a} + \frac{\cos(ax)}{a^2}\\ =& x \sin(x) + \cos(x) + C \end{align}

And that is the right answer. But I couldn't think of any problems to solve with this method that I couldn't have done with integration by parts. So I wonder if this is actually an integration technique, and whether there are problems that are best solved with that method.

$\endgroup$

2 Answers

$\begingroup$

It is a legitimate method. Here is an extension of it that is more convenient than integration by parts for evaluating the integral below

$$\int_{0}^{\infty }x^ne^{-x}dx= (-1)^n \frac{d^n}{da^n} \bigg( \int_{0}^{\infty }e^{-ax}dx\bigg)_{a=1} = (-1)^n \frac{d^n}{da^n} \frac1a\bigg|_{a=1}=n!$$

$\endgroup$ $\begingroup$

In fact, I just posted an answer with this kind on trick enter link description here where I first introduced a parameter, then differentiated, and finally reintegrated. I suppose there are more sophisticated examples around.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

You Might Also Like