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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/   
    
Associated Topics:
High School Calculus

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