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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[2] 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 

     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]/2, but there is no elementary 
antiderivative of this integrand.     

- Doctor Pete, The Math Forum 

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!
Associated Topics:
College Calculus

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