Integral of dx/(1+(tanx)^sqrt2)
Date: 06/28/2003 at 20:37:26 From: Michael Subject: A challenging integration Find the value of the integral that ranges from 0 to pi/2 of dx/(1 + (tanx)^S), where S is the square root of 2. I know this question has a solution, but I haven't the slightest idea how to do it. I tried a substitution letting z = 1 + (tanx)^S, and that eventually brought me to a rather interesting fraction after simplifying the trig functions. Nonetheless, no matter what I do, I never arrive at an answer. The exponent on the tan(x) is the square root of 2, so it is intimidating just to look at.
Date: 06/29/2003 at 09:18:05 From: Doctor Pete Subject: Re: Extremely challenging integration Hi Michael, This question is an old Putnam problem, and the solution is very well known. The reason that the exponent S = Sqrt was chosen is so that the integrand is not integrable in closed form. Thus any efforts to try to substitute or transform the integrand will fail. The key to solving this question is to use the fact that the integral is definite, and that the integrand obeys a particular property which I will describe presently. If we consider the substitution of the form u = Pi/2 - x, du = - dx, it is clear that the interval of integration is reversed and the integrand is negated, the total effect of which is that we again arrive at a definite integral over [0, Pi/2]. However, the integrand becomes du/(1 + Tan[Pi/2 - u]^S), and since Tan[Pi/2 - u] = Cot[u] = 1/Tan[u], we find the integrand is du/(1 + 1/Tan[u]^S). Now simplify the fraction to obtain Tan[u]^S du/(Tan[u]^S + 1). One should now realize that the value of the integral has not changed, but the integrand takes on a convenient form. In particular, twice the integral's value is simply 1/(1 + Tan[u]^S) + Tan[u]^S/(Tan[u]^S + 1) du, which is just 1 du. Therefore the integral evaluates to Pi/4. As you can see here, definite integrals can sometimes be evaluated by somewhat non-traditional means, where an indefinite integral would fail. This is because the integrand may satisfy certain special properties over a specific interval of integration. For instance, the integral of Log[Sin[x]] dx on [0, Pi/2] is equal to -Pi Log/2, but there is no elementary antiderivative of this integrand. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/
Date: 06/29/2003 at 13:51:08 From: Michael Subject: Thank you (A challenging integration) Thank you so much! The solution is so very easy, but I would never have thought of that in a million years. This is one of those times when you realize how beautiful math really is. Thank you so much, the answer makes perfect sense to me now!
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