Question

@Turing: Nifty integral trick that I've never seen taught before, thought you might like it

Answer 1

Thanks in advance!

Answer 2

happy new yr in advance!

Answer 3

In the mean time, I recently learned of one:\[\int_{a}^{b}\int_{c}^{d}f(x)g(y)dxdy=\int_{a}^{b}g(y)dy\cdot\int_{c}^{d}f(x)dx\]

Answer 4

really? are you sure that's good?
does it work for indefinite?

Answer 5

i asked myself the same question but unfortunately i havent seen it being applied to indefinite integrals. i would suppose that it doesnt though.

Answer 6

\[ \int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\]
Which is a commonly known result. What about
\[\int_{-\infty}^\infty x^2 e^{-ax^2}dx\]
?
You could use integration by parts, or you could say that
\[\int_{-\infty}^\infty x^2 e^{-ax^2}dx = \int_{-\infty}^\infty-\frac{\partial}{\partial a} e^{-ax^2}= -\frac{d}{d a} \int_{-\infty}^\infty e^{-ax^2}dx \]
\[= -\frac{d}{d a} \sqrt{\frac{\pi}{a}} = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]

Answer 7

@ algebra
oh no that's right it's just evaluating the area...
I know that it just looked strange for a moment.

Answer 8

Now that is very nifty...
so that how is that different than differentiation under the integral sign?

Answer 9

wow! thats interesting but i dont seem to understand how you managed to "factor out" the partial derivative (im slow).

Answer 10

It's a special case if that right?

Answer 11

Now, it might seem like that isn't much of an improvement over integration by parts, which in this case is true, but what about
\[\int_{-\infty}^\infty x^8 e^{-ax^2}dx \]
? you would have to integrate by parts four times, which isn't necessarily hard but I'd rather take the fourth derivative instead. This is more or less the same as differentiation under the integral sign, but what we're actually doing here is rewriting the integrand as a derivative as opposed to differentiating the entire thing, doing the integral, and then integrating again.

Answer 12

There is more subtlety to this trick as well. In asymptotic analysis, you sometimes find yourself rewriting the integrand as an infinite series and integrating term by term. What about
\[\int_{-\infty}^\infty \cos(x)e^{-ax^2} dx\]
?
We can rewrite that as
\[ \int_{-\infty}^\infty \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}e^{-ax^2} dx = \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_{-\infty}^\infty x^{2n} e^{-ax^2}dx\]

Answer 13

Integrating each term by parts would be a little grotesque, but:

\[ \sum_{n=0}^\infty \frac{(-1)^n}{n!} \cdot \frac{d^n}{da^n}\sqrt{\frac{\pi}{a}} = \sqrt{\pi} \cdot \sum_{n=0}^\infty \frac{(-1)^n}{n!} (-1)^n \cdot (\frac{1}{2}\cdot\frac{3}{2} \cdot \frac{5}{2} ... )\frac{1}{a^n} \]
\[= \sqrt{\pi}\cdot \sum_{n=0}^\infty \frac{1}{2^n a^n n!}\cdot(1\cdot 3 \cdot 5 \cdot 7 ... (2n-1) )\]

Answer 14

One final note is that the parameter a need not even be present to use this. You can simply insert it and then, after the problem is fully integrated, set it to one.

Answer 15

brilliant, man!
good show!

Answer 16

wow

Answer 17

@jemurray3 - what you have shown is really interesting and I want to learn more about it. do you have a link to more information on this method? or does it have a name?

Answer 18

Thank you :) It's a fun little trick that I worked out while doing integrals in quantum mechanics and thermodynamics, but I have to give most of the credit to my thermo professor for inspiring the idea. After I'd played with it awhile and worked out some kinks I found out its real name, looked it up, and refined what I knew before. Also, as a waay too late reply to pre-algebra, you can exchange the derivative and integral signs if the integrand meets certain standards of "good - behavior" and the two operations are independent (differentiation wrt a and integration wrt x, for instance). Convergent power series are always integrable term by term, which is why I didn't worry about it.

Answer 19

@ asnaseer it's a bit of a rarity as far as I can tell (which is a shame...) but it is called integration by parametric differentiation

http://en.wikipedia.org/wiki/Integration_using_parametric_derivatives

Answer 20

@jemurray3 - many thanks for the link - you sir are a star ;-)

Answer 21

There is a book called Advanced Calculus by Frederick Woods ( I think) that is now out of print but I bought it on Amazon from somebody, and it describes differentiation and integration under the equals sign in some depth.

Answer 22

Many thanks :)

Answer 23

This is much more clear than Wikipedia
Much thanks to you!

Answer 24

Whoa. not under the equals sign, under the integral sign o.0 haha

Answer 25

Just as a brief afterward, I rechecked this and I did make a few typos... that final thing should be
\[ \sqrt{\pi} \cdot \sum_{n=0}^\infty \frac{(-1)^n}{2^n a^{\frac{2n+}{2}}(2n)!}\cdot(1\cdot 2 \cdot 3 ... (2n-1)) = \sqrt{\pi} \cdot \sum_{n=0}^\infty \frac{(-1)^n (2n-1)!!}{2^n a^{\frac{2n+1}{2}}(2n)!} \]
Which turns out to be
\[ \sqrt{\frac{\pi}{a}}e^{\frac{-1}{4\sqrt{a}}} \]
According to wolfram alpha, this is in fact the value integral we're looking for. Which is nice, I suppose. Just as a final thought, I'd like to double check using a complex integral...

Answer 26

Make that
\[\sqrt{\frac{\pi}{a}} e^{\frac{-1}{4a}} \]
not my day with typos.... :)

Answer 27

\[ \int \cos(x) e^{-ax^2} dx= Re\left( \int e^{ix} e^{-ax^2}dx \right) \]
\[ \int e^{-a(x^2 - \frac{ix}{a})}= e^{\frac{-1}{4a}} \int e^{-a(x-\frac{i}{2a})^2}= \sqrt{\frac{\pi}{a}}e^{\frac{-1}{4a}} \]
Considerably simpler, but I just made up the question off the top of my head, real ones aren't so simple :)

Answer 28

OK guys - after a lot of searching I found out that this is actually just a special case of the "Leibniz integral rule" and was one of Feynman's favourite methods:

Feynman (1997, pp. 69-72) recalled seeing the method in Woods (1926) and remarked "So because I was self-taught using that book, I had peculiar methods for doing integrals," and "I used that one damn tool again and again."

That snippet is from: http://mathworld.wolfram.com/LeibnizIntegralRule.html

It is also described near the bottom of: http://www2.bc.cc.ca.us/resperic/Math6A/Lectures/ch5/3/FundamentalTheorem.htm

Some videos on this method:
1. http://www.youtube.com/watch?v=oi45rIJA744
2. http://wn.com/Integration_using_parametric_derivatives

Hope you find these useful and thanks again to @Jemurray3 for sharing such a wonderful method.