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First, I calculated the integral with the result:

$$\int \frac{1}{x^{7}+1}dx = \frac{1}{7}e^{\frac{-6i\pi}{7}}\ln\lvert{x-e^{\frac{i\pi}{7}}}\rvert+ \frac{1}{7}e^{\frac{-4i\pi}{7}}\ln\lvert{x-e^{\frac{3i\pi}{7}}}\rvert+ \frac{1}{7}e^{\frac{-2i\pi}{7}}\ln\lvert{x-e^{\frac{5i\pi}{7}}}\rvert+ \frac{1}{7}\ln\lvert{x+1}\rvert+ \frac{1}{7}e^{\frac{2i\pi}{7}}\ln\lvert{x-e^{\frac{9i\pi}{7}}}\rvert+ \frac{1}{7}e^{\frac{4i\pi}{7}}\ln\lvert{x-e^{\frac{11i\pi}{7}}}\rvert+ \frac{1}{7}e^{\frac{6i\pi}{7}}\ln\lvert{x-e^{\frac{13i\pi}{7}}}\rvert+c$$

Then, I generalised two things from it in the following:

$$x^{k}+1= \prod_{n=1}^{k}(x-{e^{\frac{(2n-1)i\pi}{k}}})$$

$$\int\frac{1}{x^{k}+1}dx = \frac{1}{k}\sum_{n=1}^{k} {e^{\frac{-(2n-1)(k-1)i\pi}{k}}}\ln\lvert{x-e^{\frac{(2n-1)i\pi}{k}}}\rvert$$

For $k=5$, I got:

$$\int \frac{1}{x^{5}+1}dx = \frac{1}{5}e^{\frac{-4i\pi}{5}}\ln\lvert{x-e^{\frac{i\pi}{5}}}\rvert+ \frac{1}{5}e^{\frac{-2i\pi}{5}}\ln\lvert{x-e^{\frac{3i\pi}{5}}}\rvert+\frac{1}{5}\ln\lvert{x+1}\rvert+\frac{1}{5}e^{\frac{2i\pi}{5}}\ln\lvert{x-e^{\frac{-3i\pi}{5}}}\rvert+ \frac{1}{5}e^{\frac{4i\pi}{5}}\ln\lvert{x-e^{\frac{-i\pi}{5}}}\rvert+c$$

Now my question: Is this all true? If so, how can I prove it, or is this enough? I'm really stuck. Thank you very much!

J.G.131
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Mordor_07
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    You can factor $x^k+1$ over the complex numbers. Every root is a power of $e^{\pi i/k}$. Then you can decompose $\frac{1}{x^k+1}$ into a sum of partial fractions and integrate term-by-term. But then you have a complex-linear combination of complex logarithms, and you need to make sure the answer is real-valued. – Matthew Leingang Jun 06 '24 at 15:26
  • @MatthewLeingang So now the proof would be that instead of using the formula, I do it again by hand and then see if it comes out the same? But how do I really know that it is the solution of the integral? Because when I equate it to Wolframalpha, it doesn't come out right or wrong. – Mordor_07 Jun 06 '24 at 15:37
  • You can check your results against this WolframAlpha evaluation. – Steven Clark Jun 06 '24 at 16:10
  • https://math.stackexchange.com/q/1999869/96384 , and many questions are linked from there. Compare also the ideas in https://math.stackexchange.com/q/2807302/96384 – Torsten Schoeneberg Jun 08 '24 at 15:12
  • @TorstenSchoeneberg The general formula: Why doesn't $\arctan(\frac{b}{a})$ apply here and if I multiply both summands of the antiderivative by x_k, do all real parts and all imaginary parts drop out? https://math.stackexchange.com/q/1999869/96384 – Mordor_07 Jun 12 '24 at 12:00
  • I don't understand your question. The fundamental misunderstanding here still seems to be that you wrongly believe the integral of $\frac{1}{z-c}$ is $\ln |z-c|$ for complex variables / numbers $c$, which it is not. Also, you should not multiply anything by $x_k$, but group together conjugates $x-c$ and $x-\bar c$, then use that $\arctan(x) = \frac{i}{2} (\log(i+x) -\log(i-x))$. – Torsten Schoeneberg Jun 12 '24 at 17:41

1 Answers1

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That doesn't look correct, already for $k=2$ because $\frac{1}{2}(-i\ln(|t-i|)+i\ln(|t+i|))=0$ for any $t\in\mathbb{R}$.

I'll treat the case $k=7$ but the following method works in general.

For $k\in[\![0;6]\!]$, write $\theta_k := \frac{(2k+1)\pi}{7}$ and $w_k:=\exp(i\theta_k)$. Notice that $\overline{w_k}=w_{6-k}$

$$X^7+1=\prod_{j=0}^6\left(X-w_j\right)$$

Therefore $$\begin{align} \frac{1}{X^7+1}&=\frac{1}{7}\sum_{j=0}^6\frac{w_j^6}{X-w_j}\\ & =\frac{1}{7}\sum_{j=0}^2\frac{\overline{w_j}}{X-w_j}+\frac{\overline{w_{6-j}}}{X-w_{6-j}}+\frac{1}{7}\frac{1}{X+1}\\ &=\frac{1}{7}\sum_{k=0}^2\frac{2\cos(\theta_k)X-2}{X^2-2\cos(\theta_k)+1}+\frac{1}{7}\frac{1}{X+1}\\ &=\frac{1}{7}\sum_{k=0}^2\cos(\theta_k)\frac{2X-2\cos(\theta_k)}{X^2-2\cos(\theta_k)X+1}+\frac{-2\sin^2(\theta_k)}{(X-\cos(\theta_k)^2+\sin(\theta_k)^2}+\frac{1}{7}\frac{1}{X+1} \end{align}$$

And $$\int\frac{dx}{x^7+1} = \frac{1}{7}\sum_{k=0}^2\cos(\theta_k)\ln(x^2-2\cos(\theta_k)x+1)-2\sin(\theta_k)\arctan\left(\frac{x-\cos(\theta_k)}{\sin(\theta_k)}\right)+\frac{1}{7}\ln(|1+x|)$$

It's not difficult to show that $$\begin{align} \sum_{n=1}^{7}e^{\frac{-(2n-1)6i\pi}{7}}\ln\lvert{x-e^{\frac{(2n-1)i\pi}{7}}}\rvert&=\sum_{n=1}^3e^{\frac{-(2n-1)6i\pi}{7}}\ln\lvert{x-e^{\frac{(2n-1)i\pi}{7}}}\rvert+e^{\frac{(2n-1)6i\pi}{7}}\ln\lvert{x-e^{\frac{-(2n-1)i\pi}{7}}}\rvert+\ln|x+1| \\ &=\sum_{n=1}^3 2\cos\left(\frac{(2n-1)i\pi}{7}\right)\ln(|x-e^{\frac{(2n-1)i\pi}{7}}|)+\ln|x+1|\\ &=\sum_{n=1}^3 2\cos\left(\frac{(2n-1)i\pi}{7}\right)\ln\left(\sqrt{\left|x^2-2\cos\left(\frac{(2n-1)i\pi}{7}\right)+1\right|}\right)+\ln|x+1|\\ &=\sum_{n=1}^3 \cos\left(\frac{(2n-1)i\pi}{7}\right)\ln\left(\left|x^2-2\cos\left(\frac{(2n-1)i\pi}{7}\right)+1\right|\right)+\ln|x+1|\\ \end{align}$$

So you're missing the $\arctan$ part of the antiderivative.

Ayoub
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  • I understand the answer, but somehow I don't either. Where exactly was the fault that it didn't work for me? I would just like to use my way. – Mordor_07 Jun 07 '24 at 15:45
  • @Mordor_07 It's not true that $\ln|x+z|$ is an antiderivative of $x\mapsto \frac{1}{x+z}$ (for $z\in\mathbb{C}\setminus\mathbb{R}$). The former is real valued and there's no reason why the later should be as well. – Ayoub Jun 07 '24 at 18:37
  • Hm, okay, that's how I've always done it in the real number range. How should I do it now in the complex number range? – Mordor_07 Jun 08 '24 at 12:35
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    @Mordor_07 Like i did ? Find the real part and the imaginary part and integrate them separately. For example, for $x\in\mathbb{R}$, $\frac{1}{x+i}=\frac{x-i}{x^2+1} $ so $\frac{1}{2}\ln(x^2+1) -i\arctan(x)$ is an antiderivative. – Ayoub Jun 08 '24 at 12:53
  • I now have the first integral: $\frac{1}{7}\cdot{e^{\frac{-6i\pi}{7}}}\cdot\int(\frac{x}{x^{2}-2\cos(\frac{\pi}{7})x+1}-\frac{{e^{\frac{-i\pi}{7}}}}{x^{2}-2\cos(\frac{\pi}{7})x+1})dx$ But how do I solve this, please? Is it even possible by hand? Or just use Wolframalpha? – Mordor_07 Jun 08 '24 at 15:09
  • Ignoring branch issues for now, an antiderivative of $f(z)=\frac{1}{z+c}$ is the complex logarithm $\mathrm{Log}(z+c)$, which basically by definition is $=\ln |z+c| + i\cdot \mathrm{Arg}(z+c)$. The argument function $\mathrm{Arg}$ boils down to an arctan; if the variable is real, $z=x$, you get $\displaystyle \int \dfrac{dx}{x+c} = \ln \left(\sqrt{(x+Re(c))^2 + Im(c)^2} \right) + i\cdot \arctan \left(\dfrac{Im(c)}{x+Re(c)}\right)$. Note how @Ayoub's $\int \dfrac{dx}{x+c} = \frac12 \ln(x^2+1) -i\arctan(x)$ is the special case $c=i$ of this, using $\arctan(1/x) = const - \arctan(x)$. – Torsten Schoeneberg Jun 08 '24 at 15:30
  • @Mordor_07 Everything can be computed by hand, that's exactly what i did in my answer. In your case, $\frac{e^{i\pi/7}}{x^2-2\cos(\frac{1}{7})x+1}$ is not the imaginary part of the first integral because $e^{i\pi/7}$ is not imaginary ! Find the real part and the imaginary part first. The real part looks like the derivative of a $\ln(u)$, the imaginary part looks like the derivative of some $\arctan(v)$. – Ayoub Jun 08 '24 at 21:01
  • @Ayoub I have now: $\frac{1}{x^{n}+1}$ integrated and after a partial fraction decomposition the integral came out, using the method from yesterday: $\frac{-1}{n}\sum_{k=1}^{n}(\int\frac{x_kx}{x^2-2x_k_rx+1} -\frac{x_k_r\ast+i x_k_i\ast}{x^2-2x_k_rx+1}dx)$ How do I solve this now? $x_k=e^\frac{(2k-1)\cdot i\pi}{n}; x_k_r=\cos{\frac{(2k-1)\pi}{n}; x_k_i=\sin {\frac{(2k-1)\pi}{n}}}$ – Mordor_07 Jun 09 '24 at 19:21
  • @TorstenSchoeneberg I've now posted the last integral here in the comments, then I'm done with it. What's the best way to solve this? – Mordor_07 Jun 09 '24 at 19:27