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Rare integration techniques/substitutions

  1. Mar 23, 2013 #1
    Does anyone know of any integration tecniques that aren't covered in calculus 1/2? For example, today I learned the Weierstrass Substitution. Are there other useful techniques?
  2. jcsd
  3. Mar 24, 2013 #2


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    The integral
    ##\int_{\pi/4}^{\pi/2} \log \log \tan x \, dx = \frac{\pi}{2} \log \left( \frac{\sqrt{2\pi}\, \Gamma(3/4)}{\Gamma(1/4)} \right)##
    requires knowledge of analytic number theory, specifically L-functions. The paper by Ilan Vardi Integrals, an Introduction to Analytic Number Theory has the proof.

    This technique was given the systematic analysis in Luis A. Medina and Victor H. Moll's paper A class of logarithmic integrals (DOI 10.1007/s11139-008-9148-7), where the authors study integrals of type ##\int_0^1 Q(x) \log \log 1/x \, dx## for some rational function Q(x).

    Another technique that doesn't get much coverage is the "method of brackets" by Gonzalez and Moss. I only have the arxiv paper: arXiv:0812.3356v1.
  4. Mar 24, 2013 #3
    Differentiation under the integral sign is something I've seen mentioned as useful several times, though I confess to not knowing it.
  5. Mar 24, 2013 #4
    Complex analysis is beautiful.

    The integral ##\displaystyle \int_{0}^{\infty}\frac{1}{x^4+1} \ dx## can be solved by partial fraction decomposition, as you might know from this thread. Or, we can use something called Cauchy's Residue Theorem. A lot of definite integrals become a lot easier if we do them indirectly with contour integrals in the complex plane.
  6. Mar 25, 2013 #5
    How would you do that integral with complex analysis?
  7. Mar 26, 2013 #6
    To solve the integral with complex analysis, one would have to use a line integral in the complex plane and use the Residue Theorem, as Mandelbroth mentioned.

    [itex]\displaystyle \int_{0}^{\infty}\frac{1}{x^4+1} \ dx[/itex]

    We could start by associating with the given real integral a related contour integral, of the form [itex]\displaystyle \int_{\Gamma}f(z) \ dz[/itex].

    We observe that [itex]2\displaystyle \int_{0}^{R}\frac{1}{x^4+1} \ dx = \displaystyle \int_{-R}^{R}\frac{1}{x^4+1} \ dx[/itex].

    So, we consider [itex]\displaystyle \int_{\Gamma}\frac{1}{z^4+1} \ dz[/itex], where [itex]\Gamma = [-R,R] \cup \Gamma(R)[/itex], with [itex]\Gamma(R)=R \ e^{it} \mid t \in [0,\pi][/itex] is a semicircular contour. Since we have [itex]f(z)=\frac{1}{z^4+1}[/itex], we can then use the Residue Theorem to evaluate the contour integral.
    Last edited: Mar 26, 2013
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