Another simple integralWhat follows is a recent generalization of the technique of computing some definite integrals discussed some time ago. The discussion here is a compilation of an article Lazy Student Integrals by Gregory Galperin and Gregory Ronsse (Math Magazine, v 81, n 2, April 2008, pp.152-154). Assume you are asked to evaluate the integral
as a function of α. You may observe that the substitution
an integral of the sort we considered previously. Thus we know that the latter integral is just plain π/4, independent of α! Being a curiosity, we'll obtain the same result in a different, but still a simple manner after the substitution
Adding the two expressions for I(α) gives
as expected. On the whole, the independence of the integral in (1) of α suggests that there may be a more to the integral than meets the eye. Indeed, the integral in (1) falls in the same class as, say, integrals
or
Do you see why one is 0 and the other 5? As a matter of fact, the two are almost immediate with a little piece of theory. Let f : [0, a] → R be any continuous function. Substitute
which becomes obvious if you note that the substitution just reflects function f in
Then adding the integrals in (3) shows that
To make the approach general we only need a way of finding more functions f that obey the symmetry condition. This too is simple. All such function can be obtained from
we g is any continuous function on [0, a] for which the denominator above does not vanish. Any function f that satisfies the symmetry condition comes in this form with
for any continuous function f defined on [0, 1] such that the denominator of the integrand above does not vanish. |Contact| |Front page| |Contents| |Algebra| |Up| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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