Integrating 1/(1+x^n)Date: 11/14/2001 at 17:09:51 From: Varghese Devassy Subject: Integration Hi, I was wondering whether there are methods to integrate functions of the form 1/(1+x^n). I have solved 1/(1+x^4) and am wondering about solutions for higher powers of x. Thank you, Varghese Devassy Date: 11/16/2001 at 04:22:36 From: Doctor Pete Subject: Re: Integration Hi, Let F[x,n] = x^n + 1. Then 1/F has an "elementary" antiderivative for rational values of n. However, let's consider the case where n is a positive integer. The idea is to express F as a product of linear factors; specifically, F[x,n] = (x - r[1])(x - r[2])...(x - r[n]), where r[1], r[2], ..., r[n] are the roots of F. Then we use partial fraction decomposition to express 1/F as the sum of terms of the form A[k]/(x - r[k]), for suitable coefficients A[k]. With this in mind, the roots of F are {Exp[I*Pi/n], ..., Exp[(2k-1)I*Pi/n], ..., Exp[(2n-1)I*Pi/n], where k takes on integer values from 1 to n. For the moment, never mind that these are complex numbers. Then we have r[k] = Exp[(2k-1)I*Pi/n]. The partial fraction decomposition step is tricky. The general term of the decomposition is A[k]/(x - r[k]), for some coefficient A[k]. To find this coefficient, we notice that cross-multiplication and substitution of the value x = r[k] gives A[k](r[k] - r[1])...(r[k] - r[k-1])(r[k] - r[k+1])...(r[k] - r[n]) = 1, where the product of linear factors on the left-hand side excludes the factor (r[k] - r[k]), which is zero. This product happens to be -n*Exp[-(2k-1)I*Pi/n], so A[k] = (-1/n)Exp[(2k-1)I*Pi/n] = -r[k]/n. (The proof of this I leave to you.) Now what? So far, we have expressed 1/F as the sum of linear factors of the form (-r[k]/n)/(x - r[k]), r[k] = Exp[(2k-1)I*Pi/n], for k = 1 to k = n. Motivated by the idea that complex conjugate pairs add and multiply to real numbers, let us call s[k] the complex conjugate to r[k]. Note that {s[1], s[2], ..., s[n]} = {r[1], r[2], ..., r[n]}; i.e., the set of s[k]'s is the same as the set of r[k]'s, because the complex roots of F come in conjugate pairs. So if we have the pair of terms (-r[k]/n)/(x - r[k]) + (-s[k]/n)/(x - s[k]), cross-multiplying gives ((-r[k]/n)(x - s[k]) + (-s[k]/n)(x - r[k]))/((x - r[k])(x - s[k])). Now, note that r[k] = Exp[(2k-1)I*Pi/n] = Cos[(2k-1)Pi/n] + I*Sin[(2k-1)Pi/n], q[k] = Exp[(1-2k)I*Pi/n] = Cos[(2k-1)Pi/n] - I*Sin[(2k-1)Pi/n], so in particular we have r[k] + q[k] = 2 Cos[(2k-1)Pi/n], r[k]q[k] = 1. Therefore, the sum of the conjugate pair terms is (-1/n)(r[k]x - 1 + q[k]x - 1)/(x^2 - (r[k]+q[k])x + 1) = (2/n)(1 - Cos[(2k-1)Pi/n]x)/(x^2 - 2 Cos[(2k-1)Pi/n]x + 1). Note that this expression is real. So it follows that 2/F[x,n] = Sum[(2/n)(1-Cos[(2k-1)Pi/n]x)/(x^2 - 2Cos[(2k-1)Pi/n]x + 1)] where the sum is taken over k = 1 to k = n. Now we can see where we're going, for from here it is a relatively simple matter to integrate this expression term by term. Let C[k] = Cos[(2k-1)Pi/n], S[k] = Sin[(2k-1)Pi/n], so the k(th) term of the integrand is (2/n)(1-C[k]x)/(x^2 - 2C[k]x + 1). Before we integrate, however, note that if C[k] = -1 (which happens when n is odd), the integrand simplifies to (2/n)/(x+1), which integrates to (-2/n)Log[x+1]. Otherwise, we separate the integrand into two parts, giving -(C[k]/n)(2x - 2C[k])/(x^2-2C[k]x+1) + (2S[k]^2/n)/(x^2-2C[k]x+1) The first term is easily integrated with the substitution u = x^2 - 2C[k]x + 1, du = (2x - 2C[k]) dx, giving -(C[k]/n)Log[x^2 - 2C[k]x + 1]. The second term is a bit harder. Completing the square in the denominator, we obtain (2S[k]^2/n)/(x^2 - 2C[k]x + C[k]^2 + S[k]^2) = (2S[k]^2/n)/((x - C[k])^2 + S[k]^2). With the substitution u = x - C[k], we obtain (2S[k]^2/n)/(u^2 + S[k]^2), which directly integrates to (2S[k]/n) ArcTan[(x-C[k])/S[k]]. Therefore, the integral of 1/F is half the sum of the above terms from k = 1 to k = n, which is (S[k]/n)ArcTan[(x-C[k])/S[k]] - (C[k]/(2n))Log[x^2-2C[k]x+1], whenever S[k] is nonzero (which implies C[k] is not -1), and (1/n)Log[x+1] when S[k] = 0 and C[k] = -1. Whew! I hope I didn't make a typing error, and I hope you understood my line of reasoning. Notice that most of the work was in expressing the integrand in a form that was readily integrable; only in the very last step did I actually integrate the function. Also note that I left a little fact unproven, and it might be a good exercise to try to prove it. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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