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I wish to decompose $\frac{1}{(1+x^2)(1+x^{2015})}$

I have $1 = (Ax+B)(1+x^{2015}) + (Cx+D)(1+x^2) = Ax^{2016} + Bx^{2015} + Cx^{3} + Dx^{2} + x(A+C) + B+D$

Doesn't this imply that $B+D = 1$, but we also have that $B=D = 0$ as well.

What did I do wrong?

I am examining the integral $\int_{0}^{\infty} \frac{dx}{(1+x^2)(1+x^{2015})}$

krlo
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    Would you like to post your complete problem may be it is related to definite Integral – juantheron Sep 24 '15 at 18:30
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    The numerator of the $1+x^{2015}$ must be a polynomial of order $2014$ – user211337 Sep 24 '15 at 18:30
  • What's the source of the problem ? This doesn't really make sense as it currently is. – Victor Sep 24 '15 at 18:31
  • I doubt you want to do that – Jean-Sébastien Sep 24 '15 at 18:34
  • By "examining", do you mean "checking for convergence", or actually evaluating it? – user170231 Sep 24 '15 at 18:40
  • @user170231 Both? I am attempting to find any interesting characteristics – krlo Sep 24 '15 at 18:41
  • @user211337 How is that possible? Algebraically, I do not think that I have done anything incorrect – krlo Sep 24 '15 at 18:41
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    This sounds like the classical complex analysis integral where one should count residues outside of a finite region instead of inside and use that the sum of residues is zero. – mickep Sep 24 '15 at 18:44
  • @krlo I believe your equation should be $1=(1+x^2)\sum_{i=0}^{2015}\left(a_ix^i\right) + (Ax+C)(1+x^{2015})$, for constants $a_i$. In general for an order $n$ denominator, your numerator is order $n-1$, which is why you have a $Ax+C$ from the $1+x^2$. Though, I recommend abandoning partial fractions for this one, unless there is a subtle trick I've missed. – user211337 Sep 24 '15 at 18:44
  • Checking for convergence is easy enough, since $\dfrac{1}{(1+x^2)(1+x^{2015})}\le\dfrac{1}{1+x^2}$. – user170231 Sep 24 '15 at 18:45
  • Have you done one complex variable? This could probably be solved with some symmetry argument and a contour integral. – mathreadler Sep 24 '15 at 18:47
  • @user211337 I recommend adding your comment as an answer. I'll make a new post under different tags if I need further assistance. – krlo Sep 24 '15 at 18:49

2 Answers2

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There is a trick: $$\begin{eqnarray*}\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}&=&\int_{0}^{1}\frac{dx}{(1+x^2)(1+x^m)}+\int_{1}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}\\&=&\int_{0}^{1}\frac{dx}{1+x^2}\left(\frac{1}{1+x^m}+\frac{1}{1+x^{-m}}\right)\\&=&\int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\color{red}{\frac{\pi}{4}}\end{eqnarray*}$$ no matter what $m$ is.

Explanation: split the integration range into $(0,1)\cup (1,+\infty)$; use the change of variable $x\mapsto\frac{1}{x}$ on the unbounded piece and recombine the two pieces in a single integral over $(0,1)$; simplify, simplify more.

Jack D'Aurizio
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    Wow that is neat! Thank you! Question though, how did you go from $\int_{0}^{1} \frac{dx}{(1+x^2)(1+x^m)} + \int_{1}^{\inf} \frac{dx}{(1+x^2)(1+x^m)}$ to $\int_{0}^{1} \frac{dx}{1+x^2} + \frac{1}{1+x^{-m}}$ ? – krlo Sep 24 '15 at 18:50
  • Cool trick. Could you clarify how $1/(1+x^m) + 1/(1+x^{-m}) = 1$ or what happens between row 2 and 3? – mathreadler Sep 24 '15 at 18:54
  • Er, my comment has a typo, but you posted the explanation nonetheless. I don't immediately understand since I have applied change-of-variable method in quite some time, but i'l figure it out! – krlo Sep 24 '15 at 19:11
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    @mathreadler $$\frac{1}{1+x^m} + \frac{1}{1+x^{-m}} = \frac{1}{1+x^m} + \frac{x^m}{x^m+1} = \frac{1+x^m}{1+x^m}=1$$ – Théophile Sep 24 '15 at 19:20
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So be it, but we'll use complex analysis

$$x^2+1=(x+i) \cdot (x-i)$$

The solution of $1+x^{2015}=0$ can be handled with Demoivre's Theorem. The theorem yields,

$$1+x^{2015}=\prod_{n=0}^{2014} \left(x-\cos \left({{(2\cdot n+1) \cdot \pi} \over n}\right)-i\cdot \sin\left({{(2\cdot n+1) \cdot \pi} \over n}\right) \right)=\prod_{n=0}^{2014} x-\lambda_n$$

From here, we must employ Heaviside's Cover-up Method. I'd suggest a CAS system at this point. If you like, here's the formula for the terms in the partial fraction expansion,

$${1 \over {(1+x^2) \cdot (1+x^{2015})}}={1 \over {(2i) \cdot \prod_{n=0}^{2014} (i-\lambda_n)}}+{1 \over {(2i) \cdot \prod_{n=0}^{2014} (i+\lambda_n)}}+\sum_{k=0}^{2014} \lim_{x \to {\lambda_k}} {{\lambda_k-\lambda_n} \over {(1+\lambda_k^2) \cdot \prod_{n=0}^{2014} (\lambda_k-\lambda_n)}}$$

If you'd like to evaluate the integral, this answer will suffice. He obtains $\pi/4$ using a general principle that applies to an entire class of integrals with integrands with,

$$f(x)+f(a+b-x)=k$$

Where, $a$ and $b$ are integration bounds, $k$ is a constant, and the substitution, $\tan(\theta) \rightarrow x$ should be used.

Community
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Zach466920
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  • "and the substitution, x→tan(θ) must be used." No, this substitution is useful to compute some θ-integrals by reducing them to some x-integrals, not the other way around. Please read more carefully the answer you cite. – Did Sep 25 '15 at 17:40