anonymous
  • anonymous
@Turing: Nifty integral trick that I've never seen taught before, thought you might like it
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
Thanks in advance!
EarthCitizen
  • EarthCitizen
happy new yr in advance!
anonymous
  • anonymous
In the mean time, I recently learned of one:\[\int_{a}^{b}\int_{c}^{d}f(x)g(y)dxdy=\int_{a}^{b}g(y)dy\cdot\int_{c}^{d}f(x)dx\]

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TuringTest
  • TuringTest
really? are you sure that's good? does it work for indefinite?
anonymous
  • anonymous
i asked myself the same question but unfortunately i havent seen it being applied to indefinite integrals. i would suppose that it doesnt though.
anonymous
  • anonymous
\[ \int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\] 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 \[\int_{-\infty}^\infty x^2 e^{-ax^2}dx = \int_{-\infty}^\infty-\frac{\partial}{\partial a} e^{-ax^2}= -\frac{d}{d a} \int_{-\infty}^\infty e^{-ax^2}dx \] \[= -\frac{d}{d a} \sqrt{\frac{\pi}{a}} = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]
TuringTest
  • TuringTest
@ algebra oh no that's right it's just evaluating the area... I know that it just looked strange for a moment.
TuringTest
  • TuringTest
Now that is very nifty... so that how is that different than differentiation under the integral sign?
anonymous
  • anonymous
wow! thats interesting but i dont seem to understand how you managed to "factor out" the partial derivative (im slow).
TuringTest
  • TuringTest
It's a special case if that right?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[ \int_{-\infty}^\infty \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}e^{-ax^2} dx = \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_{-\infty}^\infty x^{2n} e^{-ax^2}dx\]
anonymous
  • anonymous
Integrating each term by parts would be a little grotesque, but: \[ \sum_{n=0}^\infty \frac{(-1)^n}{n!} \cdot \frac{d^n}{da^n}\sqrt{\frac{\pi}{a}} = \sqrt{\pi} \cdot \sum_{n=0}^\infty \frac{(-1)^n}{n!} (-1)^n \cdot (\frac{1}{2}\cdot\frac{3}{2} \cdot \frac{5}{2} ... )\frac{1}{a^n} \] \[= \sqrt{\pi}\cdot \sum_{n=0}^\infty \frac{1}{2^n a^n n!}\cdot(1\cdot 3 \cdot 5 \cdot 7 ... (2n-1) )\]
anonymous
  • anonymous
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.
TuringTest
  • TuringTest
brilliant, man! good show!
anonymous
  • anonymous
wow
asnaseer
  • asnaseer
@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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
@ 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
asnaseer
  • asnaseer
@jemurray3 - many thanks for the link - you sir are a star ;-)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Many thanks :)
TuringTest
  • TuringTest
This is much more clear than Wikipedia Much thanks to you!
anonymous
  • anonymous
Whoa. not under the equals sign, under the integral sign o.0 haha
anonymous
  • anonymous
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...
anonymous
  • anonymous
Make that \[\sqrt{\frac{\pi}{a}} e^{\frac{-1}{4a}} \] not my day with typos.... :)
anonymous
  • anonymous
\[ \int \cos(x) e^{-ax^2} dx= Re\left( \int e^{ix} e^{-ax^2}dx \right) \] \[ \int e^{-a(x^2 - \frac{ix}{a})}= e^{\frac{-1}{4a}} \int e^{-a(x-\frac{i}{2a})^2}= \sqrt{\frac{\pi}{a}}e^{\frac{-1}{4a}} \] Considerably simpler, but I just made up the question off the top of my head, real ones aren't so simple :)
asnaseer
  • asnaseer
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.

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