Monday Math 28

Let us consider a function of two real variables α and β; such that is defined and continuous for all α,β>0; and which obeys, for αβ,
Does such a function exist, and if so, what is it?

Note that this exercise has three parts: First, to show that the integrand is valid over the infinite interval (0,∞); second, that the discontinuity in on the line α=β is removable; and third, to evaluate the integral, and thus find the function.

First, we note that the integrand has a singularity at the origin. However, as α,β>0, we see that and for all x≠0, and so and for all x≠0. We see further that as , , and similarly for . Thus the integrand has a removable singularity at x=0.
(In fact, the function not only has a removable singularity at the origin, with g(0)=0, but the derivatives of all order do so as well, with all order derivatives being zero at the origin; the function is “flatter” at the origin than x to any finite power.)

We also note that , and similarly for , so , which is a necessary (but not sufficient) condition for the integral to converge.
(Note also that the limit means that the integrals and each diverge, so we cannot consider them separately, as was done with the first integral in this post, which looks similar at first glance.)

Now, we tackle the second point, the problem of α=β. Let us move the constant inside our integral, and consider the resulting integrand . We recall the Maclaurin series holds for all x, and so substituting in for the variable, we find that for x≠0,
and similarly for β:
Note that the first term of both these series is 1, so when we take the difference, these terms will cancel, and we can consider only the n≥1 terms, and so:
We see that is divisible by for all n≥1, as we recall that , and so:
, thus:
, which is continuous (in α and β) and defined for α=β.
(For α=β=a, the sum becomes

Now, we only need to find the integral . First, we make the substitution , giving:
Now, here is the key insight: we know the definite integral . Note that letting , , and , the right-hand side of the above becomes our integrand, and so our integral becomes:
a double integral over a half-infinite rectangular strip. We can reverse the order of integration, getting:
The Gaussian integral , and so , and so:
Lastly, we use that for positive α and β,

to see that:
and we have our solution.


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One Response to “Monday Math 28”

  1. Dejan Stamenkovic Says:

    this is some kind of oddly parabola?

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