# Integration under the integral sign

1. Aug 27, 2011

### 54stickers

I was reading along in the book Advanced Calculus by Frederick S. Woods today and came upon this technique. I played with it a little on some sample problems, but I still cannot figure out what it has any relation to.

My questions I guess are that,
(1) what does this method help to accomplish
(2) Do I have to differentiate the final result to have a "real" answer
(3) Why is the end result seem only related to the beginning by an integration operation

I may just be completely lost, and all these things be trivial. but any help would be great, thanks!

Here is a link to an example http://mathworld.wolfram.com/IntegrationUndertheIntegralSign.html" [Broken]

Last edited by a moderator: May 5, 2017
2. Aug 28, 2011

### lurflurf

(3)The problem is worked backwards (6) is the answer.
(2)This method could be used in different ways. As used in the link you do not have to differentiate the final result to have a "real" answer.
(1)This method could be used in different ways. As used in the link one reduces an integral to an easier one.

The idea of your example is
Problem: Compute an integral
1)The integral is hard now what?
Write the integral as two integrals iterated.
2)Now there are two integrals How did that help?
In some cases the integrals are easy in revese order and give the same answer
3)Reverse order of integrals (and justify to desired level of rigor).
This integral is easy
4)Do easy integral.
Done, rejoice.

Last edited: Aug 28, 2011
3. Aug 28, 2011

### 54stickers

That makes sense, thanks.

4. Sep 1, 2011

### 54stickers

Sorry for bump and double post,

but I seem to still not understand after doing a few more examples, and trying to "Differentiate under the integral" of the integral in (6). That was quite futile for me.

In that link, is the Integral to be solved (6)?

If yes, how was the equation (2) deduced from (6)?

Thanks for the help!

5. Sep 1, 2011

### Mute

Perhaps the example page could be written more clearly. Evaluating the integral in (6) is the goal of that work. The trick in that example is to recognize that

$$\frac{x^b-x^a}{\ln x} = \int_a^b d\alpha x^\alpha.$$

So, if you want to calculate

$$\int_0^1 dx~\frac{x^b-x^a}{\ln x},$$
you can start by replacing with integrand with the equivalent integral. Then, you can interchange the order of the integrals (by Fubini's theorem, assuming it applies), and it turns out that the integral $\int_0^1 dx~x^\alpha$ is easier to do than the original integral, and is just equal to $1/(\alpha+1)$ (for $\alpha \neq -1$).

So, you've reduced the original problem - evaulating the integral in (6) - to evaluating

$$\int_a^b d\alpha~\frac{1}{\alpha+1}.$$

So, to summarize:

If you have a hard integral to do, but you recognize your integrand as being equivalent to another integral, you can replace the integrand with the equivalent integral, and then using Fubini's theorem you can change the order of integration. The inner integral is hopefully relatively easy to do, so you can perform that integral, and then hopefully you end up with a new integrand that's easier to integrate than the original integrand.

6. Sep 2, 2011

### 54stickers

That's quite a lot easier to follow!

Thanks!

Last edited: Sep 2, 2011