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W. W. Hansen's Trick to Evaluating Integrals

  1. Mar 2, 2012 #1


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    In "Probability Theory: The Logic of Science", the author E. T. Jaynes relates that Prof. Hansen at Stanford evaluated integrals by treating constants like pi in an integrand as a variable. Sounds fantastic! Does anyone know how this is done?
  2. jcsd
  3. Mar 5, 2012 #2
    I am going out on a limb here, and this is probably wrong. But the one thing that comes to my mind is the fact that
    [tex] \int_{0}^{a} f(x) \mathrm{d}x = \int_{0}^{a} f(a-x) \mathrm{d}x[/tex]
    For an concerete example, let us evaluate
    [tex] \int_{0}^{\pi/2} \frac{\sin x}{\cos x + \sin x} \mathrm{d}x [/tex]
    I am sure you can fill in the rest, using the hint above =)
    If this is not the trick, I am very interested in it as well!
  4. Mar 5, 2012 #3


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    I'm guessing the trick amounts to replacing a constant in the integrand (or integration limits) with a variable, which allows you to then differentiate (integrate) with respect to that variable to make the integral easier to do. At the end of the calculation you integrate (differentiate) to get the original result you wanted. The classic example is evaluating Gaussian integrals. Consider

    [tex]\int_{-\infty}^\infty dx~x^2 \exp\left(-x^2\right)[/tex]

    You could try substitution to evaluate this, but if you know the result [itex]\int_{-\infty}^\infty dx~\exp\left(-ax^2\right) = \sqrt{\pi/a}[/itex] you can easily solve the integral as follows: first, insert a variable 'a' into the exponential, then note

    [tex]\int_{-\infty}^\infty dx~x^2 \exp\left(-a x^2\right) = -\frac{\partial }{\partial a}\left(\int_{-\infty}^\infty dx~\exp\left(-a x^2\right)\right) = -\frac{\partial }{\partial a}\left(\frac{\pi}{a}\right)^{1/2} = \frac{1}{2}\frac{\sqrt{\pi}}{a^{3/2}}[/tex]

    Setting a = 1 gives the result of the original integral.
  5. Mar 6, 2012 #4


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    Neb, I don't see how what you wrote is at all useful...

    Mute, now that's another story! This is very clever.

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