6 ms·

An integral with a couple lessons

61 points by douche 10y ago

klodolph 10y ago
Only in a certain sense does that antiderivative not exist.
- kmill 10y ago
  Do you mean something other than the sense the author mentions, that it can't be given in terms of elementary functions?
  - oxymoron 10y ago
    Can't be given in terms of elementary functions, with a finite number of terms. Seems a bit arbitrary though -- neither can exp(x) or ln(x) or sin(x), so introducing erf(x) and defining it using an integral seems more par the course than cheating.
    - kmill 10y ago
      Yeah, I meant finitely many terms. To me, things like "infinite sums" are special operations since they don't always converge. In case and exp, ln, sin seem special: they are solutions to simple linear differential equations x'-x=0, tx'=1, and x''-x=0. Though, erf comes from x''+2tx'=0, so I can't say I really understand what makes a function "elementary." Without knowing more, I'd exclude logarithms, and then say that elementary functions are solutions to homogeneous linear differential equations with constant coefficients. Edit: looking at MathWorld, it seems the definition has the set of elementary functions closed under inverses, which is why logarithms are included.
      - OscarCunningham 10y ago
        I don't know a precise definition of "elementary", but one thing about the elementary functions (exp,log,sin,cos,arcsin,arccos,+,x,-,/) is that they interact with each other in nice ways: log(ab)=log(a)+log(b) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) sin(log(x))=(x^-i - x^i)i/2 etc. This means that if your integral yields and answer in terms of log, exp or trigonometric functions then there is a good chance that any further algebra you have to do will work out nicely. But erf satisfies no such relations and so an answer in terms of it is often an algebraic dead end.
    - qntty 10y ago
      exp, ln and sin are elementary functions, so they can be given in terms of elementary functions.
  - rwj 10y ago
    There is a specific definition for elementary functions (see Wikipedia). However, the elementary functions are only a subset of all possible functions.
    - kmill 10y ago
      What you are saying is completely true, but I don't understand what you are meaning to address. The person I was responding to (klodolph) seemed to be correcting the article's author about something which needed no correction.
tomahunt 10y ago
Another example of the main point in this post is a simpler integral 0 to 2*pi of (sin x)^2 or (cos x)^2. By noting that both integrals will be the same and looking at the integral of their sum. The Gaussian integral in the post is on of the most useful out there. I highly recommend the Wikipedia page: https://en.m.wikipedia.org/wiki/Gaussian_integral https://en.m.wikipedia.org/wiki/Gaussian_integral Quantum Field Theory in a Nutshell by Zee bases a couple of chapters on this integral.
- n4r9 10y ago
  When I was preparing for the UK STEP examinations, one of the questions had you evaluate an integral I by carefully choosing another integral J such that I+J and I-J were both very easy to evaluate. From there it's just a matter of simultaneous equations. I've never seen this trick replicated elsewhere in research or teaching, but it's stuck with me as an incredibly elegant method.
taneq 10y ago
When I saw the title I immediately wondered if it was the integral of e^-(x^2). I spent months in high school wrangling with this one, convinced that it had to be possible somehow. Obviously I never quite made it happen, but some good times were had nonetheless.
ohazi 10y ago
It's been ages since I took complex analysis, but I distinctly remember the first time I used contour integration to solve a convoluted definite integral and being absolutely floored that I got the right answer. The technique looks like super hand-wavey bullshit, and the algebra usually ends up being way easier than you think is reasonable for a given problem. I'm still amazed that it works. Example: https://en.wikipedia.org/wiki/Residue_theorem https://en.wikipedia.org/wiki/Residue_theorem
- chestervonwinch 10y ago
  For the integral in the post, it's not so straight-forward to use complex contour integration [1]. Though, I agree with you -- for many other similarly intimidating integrals, complex integration can work wonders. [1]: http://www.jstor.org/stable/2588989?seq=1#page_scan_tab_contents http://www.jstor.org/stable/2588989?seq=1#page_scan_tab_cont...
- vardhanw 10y ago
  I recently came across a beautiful application of it to solve one of Ramanujan's integrals [1]. [1] http://math.stackexchange.com/questions/808452/proof-of-a-ramanujan-integral http://math.stackexchange.com/questions/808452/proof-of-a-ra...
wiz21c 10y ago
To me the polar coordinate trick show why maths are hard for me. When I've seen this integral for the first time, I was indeed clueless. But then there's this trick. The problem is that, when I look at the integral, with the math I know when I encounter it for the first time, it's very hard to infer that the trick is possible by just looking at the integral. But maybe I don't know enough math to have the "bag of tricks" ready in my mind.
- jgrahamc 10y ago
  I did quite a lot of mathematics and your last line is correct. There are lots of hard concepts in maths that you end up getting your head around but then you end up with a toolbox of things to work with.
  - k__ 10y ago
    This often eluded me in university. I had the feeling before uni, these "tools" were shortcuts for a bit more complicated processes. But the magnitude of "complicated" got risen rather high in uni. In highschool I could simply cobble together the solution while working on it at an exam, but this "cobbling" often would take a few days for higher math. Math went from a "get rules"-subject to a "memorize all the shortcuts"-subject So university math was suddendly the same as highschool history, which I hated, and less like highschool math, which I loved.
    - pif 10y ago
      > Math went from a "get rules"-subject to a "memorize all the shortcuts"-subject I'm sincerely sorry for you, you clearly had no luck in the choice of uni/prof. It must have been painful.
      - k__ 10y ago
        It was. It's a miracle that I got through it somehow. Calculus was especially hard. Linear Algebra was rather nice.
    - 10y ago
yummyfajitas 10y ago
One very practical place I've seen where this is a problem is in monte carlo simulation. A lot of people think of the problem in terms of the monte carlo simulation. As a result, they forget that the monte carlo simulation is actually an approximation of an integral. If you remember this fact, there's a lot of tricks you can do. You can analytically evaluate parts of the integral. You can replace monte carlo with a more accurate quadrature rule; e.g. Simpson's rule in low dimensions, quasimonte carlo in high. But you'll never do any of that if you don't distinguish between the integration technique (anti-derivative trick, monte carlo, etc) and the integral itself.
todd8 10y ago
For those interested in math and particularly in learning how to attack challenging (definite) integrals, I can recommend this book by Paul Nahin: Inside Interesting Integrals [1]. It's well reviewed on Amazon and I enjoyed the writing. [1] https://www.amazon.com/gp/product/1493912763/ref=oh_aui_search_detailpage?ie=UTF8&psc=1 https://www.amazon.com/gp/product/1493912763/ref=oh_aui_sear...

61 points by douche 10y ago

klodolph 10y ago

Only in a certain sense does that antiderivative not exist.

kmill 10y ago
Do you mean something other than the sense the author mentions, that it can't be given in terms of elementary functions?
- oxymoron 10y ago
  Can't be given in terms of elementary functions, with a finite number of terms. Seems a bit arbitrary though -- neither can exp(x) or ln(x) or sin(x), so introducing erf(x) and defining it using an integral seems more par the course than cheating.
  - kmill 10y ago
    Yeah, I meant finitely many terms. To me, things like "infinite sums" are special operations since they don't always converge. In case and exp, ln, sin seem special: they are solutions to simple linear differential equations x'-x=0, tx'=1, and x''-x=0. Though, erf comes from x''+2tx'=0, so I can't say I really understand what makes a function "elementary." Without knowing more, I'd exclude logarithms, and then say that elementary functions are solutions to homogeneous linear differential equations with constant coefficients. Edit: looking at MathWorld, it seems the definition has the set of elementary functions closed under inverses, which is why logarithms are included.
    - OscarCunningham 10y ago
      I don't know a precise definition of "elementary", but one thing about the elementary functions (exp,log,sin,cos,arcsin,arccos,+,x,-,/) is that they interact with each other in nice ways: log(ab)=log(a)+log(b) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) sin(log(x))=(x^-i - x^i)i/2 etc. This means that if your integral yields and answer in terms of log, exp or trigonometric functions then there is a good chance that any further algebra you have to do will work out nicely. But erf satisfies no such relations and so an answer in terms of it is often an algebraic dead end.
  - qntty 10y ago
    exp, ln and sin are elementary functions, so they can be given in terms of elementary functions.
- rwj 10y ago
  There is a specific definition for elementary functions (see Wikipedia). However, the elementary functions are only a subset of all possible functions.
  - kmill 10y ago
    What you are saying is completely true, but I don't understand what you are meaning to address. The person I was responding to (klodolph) seemed to be correcting the article's author about something which needed no correction.

tomahunt 10y ago

Another example of the main point in this post is a simpler integral 0 to 2*pi of (sin x)^2 or (cos x)^2. By noting that both integrals will be the same and looking at the integral of their sum. The Gaussian integral in the post is on of the most useful out there. I highly recommend the Wikipedia page: https://en.m.wikipedia.org/wiki/Gaussian_integral https://en.m.wikipedia.org/wiki/Gaussian_integral Quantum Field Theory in a Nutshell by Zee bases a couple of chapters on this integral.

n4r9 10y ago
When I was preparing for the UK STEP examinations, one of the questions had you evaluate an integral I by carefully choosing another integral J such that I+J and I-J were both very easy to evaluate. From there it's just a matter of simultaneous equations. I've never seen this trick replicated elsewhere in research or teaching, but it's stuck with me as an incredibly elegant method.

taneq 10y ago

When I saw the title I immediately wondered if it was the integral of e^-(x^2). I spent months in high school wrangling with this one, convinced that it had to be possible somehow. Obviously I never quite made it happen, but some good times were had nonetheless.

ohazi 10y ago

It's been ages since I took complex analysis, but I distinctly remember the first time I used contour integration to solve a convoluted definite integral and being absolutely floored that I got the right answer. The technique looks like super hand-wavey bullshit, and the algebra usually ends up being way easier than you think is reasonable for a given problem. I'm still amazed that it works. Example: https://en.wikipedia.org/wiki/Residue_theorem https://en.wikipedia.org/wiki/Residue_theorem

chestervonwinch 10y ago
For the integral in the post, it's not so straight-forward to use complex contour integration [1]. Though, I agree with you -- for many other similarly intimidating integrals, complex integration can work wonders. [1]: http://www.jstor.org/stable/2588989?seq=1#page_scan_tab_contents http://www.jstor.org/stable/2588989?seq=1#page_scan_tab_cont...
vardhanw 10y ago
I recently came across a beautiful application of it to solve one of Ramanujan's integrals [1]. [1] http://math.stackexchange.com/questions/808452/proof-of-a-ramanujan-integral http://math.stackexchange.com/questions/808452/proof-of-a-ra...

wiz21c 10y ago

To me the polar coordinate trick show why maths are hard for me. When I've seen this integral for the first time, I was indeed clueless. But then there's this trick. The problem is that, when I look at the integral, with the math I know when I encounter it for the first time, it's very hard to infer that the trick is possible by just looking at the integral. But maybe I don't know enough math to have the "bag of tricks" ready in my mind.

jgrahamc 10y ago
I did quite a lot of mathematics and your last line is correct. There are lots of hard concepts in maths that you end up getting your head around but then you end up with a toolbox of things to work with.
- k__ 10y ago
  This often eluded me in university. I had the feeling before uni, these "tools" were shortcuts for a bit more complicated processes. But the magnitude of "complicated" got risen rather high in uni. In highschool I could simply cobble together the solution while working on it at an exam, but this "cobbling" often would take a few days for higher math. Math went from a "get rules"-subject to a "memorize all the shortcuts"-subject So university math was suddendly the same as highschool history, which I hated, and less like highschool math, which I loved.
  - pif 10y ago
    > Math went from a "get rules"-subject to a "memorize all the shortcuts"-subject I'm sincerely sorry for you, you clearly had no luck in the choice of uni/prof. It must have been painful.
    - k__ 10y ago
      It was. It's a miracle that I got through it somehow. Calculus was especially hard. Linear Algebra was rather nice.
  - 10y ago

yummyfajitas 10y ago

One very practical place I've seen where this is a problem is in monte carlo simulation. A lot of people think of the problem in terms of the monte carlo simulation. As a result, they forget that the monte carlo simulation is actually an approximation of an integral. If you remember this fact, there's a lot of tricks you can do. You can analytically evaluate parts of the integral. You can replace monte carlo with a more accurate quadrature rule; e.g. Simpson's rule in low dimensions, quasimonte carlo in high. But you'll never do any of that if you don't distinguish between the integration technique (anti-derivative trick, monte carlo, etc) and the integral itself.

todd8 10y ago

For those interested in math and particularly in learning how to attack challenging (definite) integrals, I can recommend this book by Paul Nahin: Inside Interesting Integrals [1]. It's well reviewed on Amazon and I enjoyed the writing. [1] https://www.amazon.com/gp/product/1493912763/ref=oh_aui_search_detailpage?ie=UTF8&psc=1 https://www.amazon.com/gp/product/1493912763/ref=oh_aui_sear...