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Jemurray3

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    Thanks in advance!

  2. EarthCitizen
    • 3 years ago
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    happy new yr in advance!

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

  4. TuringTest
    • 3 years ago
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    really? are you sure that's good? does it work for indefinite?

  5. pre-algebra
    • 3 years ago
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    i asked myself the same question but unfortunately i havent seen it being applied to indefinite integrals. i would suppose that it doesnt though.

  6. Jemurray3
    • 3 years ago
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    \[ \int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\] 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 \[\int_{-\infty}^\infty x^2 e^{-ax^2}dx = \int_{-\infty}^\infty-\frac{\partial}{\partial a} e^{-ax^2}= -\frac{d}{d a} \int_{-\infty}^\infty e^{-ax^2}dx \] \[= -\frac{d}{d a} \sqrt{\frac{\pi}{a}} = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]

  7. TuringTest
    • 3 years ago
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    @ algebra oh no that's right it's just evaluating the area... I know that it just looked strange for a moment.

  8. TuringTest
    • 3 years ago
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    Now that is very nifty... so that how is that different than differentiation under the integral sign?

  9. pre-algebra
    • 3 years ago
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    wow! thats interesting but i dont seem to understand how you managed to "factor out" the partial derivative (im slow).

  10. TuringTest
    • 3 years ago
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    It's a special case if that right?

  11. Jemurray3
    • 3 years ago
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    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.

  12. Jemurray3
    • 3 years ago
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    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 \[ \int_{-\infty}^\infty \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}e^{-ax^2} dx = \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_{-\infty}^\infty x^{2n} e^{-ax^2}dx\]

  13. Jemurray3
    • 3 years ago
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    Integrating each term by parts would be a little grotesque, but: \[ \sum_{n=0}^\infty \frac{(-1)^n}{n!} \cdot \frac{d^n}{da^n}\sqrt{\frac{\pi}{a}} = \sqrt{\pi} \cdot \sum_{n=0}^\infty \frac{(-1)^n}{n!} (-1)^n \cdot (\frac{1}{2}\cdot\frac{3}{2} \cdot \frac{5}{2} ... )\frac{1}{a^n} \] \[= \sqrt{\pi}\cdot \sum_{n=0}^\infty \frac{1}{2^n a^n n!}\cdot(1\cdot 3 \cdot 5 \cdot 7 ... (2n-1) )\]

  14. Jemurray3
    • 3 years ago
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    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.

  15. TuringTest
    • 3 years ago
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    brilliant, man! good show!

  16. meverett04
    • 3 years ago
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    wow

  17. asnaseer
    • 3 years ago
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    @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?

  18. Jemurray3
    • 3 years ago
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    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.

  19. Jemurray3
    • 3 years ago
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    @ 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

  20. asnaseer
    • 3 years ago
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    @jemurray3 - many thanks for the link - you sir are a star ;-)

  21. Jemurray3
    • 3 years ago
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    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.

  22. Jemurray3
    • 3 years ago
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    Many thanks :)

  23. TuringTest
    • 3 years ago
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    This is much more clear than Wikipedia Much thanks to you!

  24. Jemurray3
    • 3 years ago
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    Whoa. not under the equals sign, under the integral sign o.0 haha

  25. Jemurray3
    • 3 years ago
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    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...

  26. Jemurray3
    • 3 years ago
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    Make that \[\sqrt{\frac{\pi}{a}} e^{\frac{-1}{4a}} \] not my day with typos.... :)

  27. Jemurray3
    • 3 years ago
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    \[ \int \cos(x) e^{-ax^2} dx= Re\left( \int e^{ix} e^{-ax^2}dx \right) \] \[ \int e^{-a(x^2 - \frac{ix}{a})}= e^{\frac{-1}{4a}} \int e^{-a(x-\frac{i}{2a})^2}= \sqrt{\frac{\pi}{a}}e^{\frac{-1}{4a}} \] Considerably simpler, but I just made up the question off the top of my head, real ones aren't so simple :)

  28. asnaseer
    • 3 years ago
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    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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