What 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 x = tanθ leads to

(1)

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 u = 1/x. This substitution yields

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

(2)

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 u = x - a to obtain

(3)

which becomes obvious if you note that the substitution just reflects function f  in x = a/2. Now suppose that f satisfies the following symmetry condition

f(x) + f(a - x) = 1.

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 g = f ! In particular, (2) is obtained with g(x) = 3 + x^√7 while (1) follows from sin(x) = cos(π/2 - x). But we see that even more generally

for any continuous function f defined on [0, 1] such that the denominator of the integrand above does not vanish.

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