$$\int\sqrt[4] {\tan \left( x\right) } dx$$
I'm really stuck right now with this integral, so any kind of advice would be appreciated.
Perhaps a pretty nifty substitution that can save me from partial fraction decomposition.
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1it would help if you typeset it correctly, in the body of the question – Will Jagy Apr 11 '16 at 00:46
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1If you make the substitution $z^4=tanx$ you get $4z^3dz=sec^2 x dx = (z^8+1)dx\implies dx = \frac {4z^3}{z^8+1} dz$ Not pleasant... – lulu Apr 11 '16 at 00:49
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I also think is the logical substitution (you want to get rid of the square root) but how can I do the partial fraction decomposition with that horrible $z^{8}+1$ – martianfever Apr 11 '16 at 01:06
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It's just roots of unity. That said, I don't envy you the task. – lulu Apr 11 '16 at 01:36
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It is quite easy, you solve for the roots of $z^8 = -1$ in the complex plane. If we denote the roots by $\alpha_n$, then the partial fraction decomposition is $\sum_{n=1}^{8}\frac{A_n}{z-\alpha_n}$, and we have $A_n = 4\alpha_n^3\lim_{z\to\alpha_n}\frac{z-\alpha_n}{z^8+1}$, which is trivial to evaluate using L'Hopital's rule. You then get a summation over complex logarithms, which you can easily rewrite in terms of real logarithms and arctan functions. After all $\log(a+ b i)$ can be written as $\log(r \exp(i \theta))$ with $r = \sqrt(a^2+b^2)$ and $\theta = \arctan(\frac{b}{a})$ – Count Iblis Apr 11 '16 at 01:36
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$$ \small \left( x^2 + \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 + \sqrt 2} \; x + 1 \right) \left( x^2 + \sqrt{2 - \sqrt 2} \; x + 1 \right) \left( x^2 - \sqrt{2 - \sqrt 2} \; x + 1 \right) = x^8 + 1 $$
Will Jagy
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I usually factor a cyclotomic denominator like this into linear factors, expand into partial fractions, integrate to logarithms, and then convert those complex logarithms into logarithms and arctangents, but this time I thought I would try combining complex conjugate partial fractions before integrating and see how it went. First the substitution $z^4=\tan x$: $$\int\sqrt[4]{\tan x}dx=4\int\frac{z^4}{z^8+1}dz$$ And now $$\frac{4z^4}{z^8+1}=\frac{4z^4}{\prod_{k=0}^7(z-\omega_k)}=\sum_{k=0}^7\frac{A_k}{z-\omega_k}$$ Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, and $\theta_k=\frac{(2k+1)\pi}8$ And then as @Count Iblis remarked, it's easy to get the numerators by L'Hopital's rule $$\lim_{z\rightarrow\omega_k}\frac{4z^4}{z^8+1}=\lim_{z\rightarrow\omega_k}\frac{4z^4}{8z^7}=\frac1{2\omega_k^3}=-\frac12\omega_k^5$$ Combining complex conjugate fractions, $$\begin{align}\frac{4z^4}{z^8+1} & =\sum_{k=0}^7\frac{-\frac12\omega_k^5}{z-\omega_k}=\sum_{k=0}^3\frac{-\frac12\omega_k^5(z-\omega_{7-k})-\frac12\omega_{7-k}^5(z-\omega_k)}{(z-\omega_k)(z-\omega_{7-k})}\\ & =\sum_{k=0}^3\frac{-z\cos5\theta_k}{z^2-2z\cos\theta_k+1}=\sum_{k=0}^3\frac{(z-\cos\theta_k)\cos3\theta_k+\cos\theta_k\cos3\theta_k}{z^2-2z\cos\theta_k+1}\end{align}$$ So $$\begin{align}\int\sqrt[4]{\tan x}dx & =\sum_{k=0}^3\int\frac{(z-\cos\theta_k)\cos3\theta_k+\cos\theta_k\cos3\theta_k}{z^2-2z\cos\theta_k+1}d\theta\\ & =\sum_{k=0}^3\left[\frac12\cos3\theta_k\ln\left(z^2-2z\cos\theta_k+1\right)+\frac{\cos3\theta_k\cos\theta_k}{\sin\theta_k}\tan^{-1}\left(\frac{z-\cos\theta_k}{\sin\theta_k}\right)\right]+C\\ & =\sum_{k=0}^3\left[\frac12\cos3\theta_k\ln\left(\sqrt{\tan x}-2\sqrt[4]{\tan x}\cos\theta_k+1\right)+\frac{\cos3\theta_k\cos\theta_k}{\sin\theta_k}\tan^{-1}\left(\frac{\sqrt[4]{\tan x}-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$ Evaluating the trig functions, $$\begin{align}\int\sqrt[4]{\tan x}dx & =\frac14\sqrt{2-\sqrt2}\ln\left(\sqrt{\tan x}-\sqrt[4]{\tan x}\sqrt{2+\sqrt2}+1\right)+\\ & \frac12\sqrt{2+\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4+2\sqrt2}-\frac{2+\sqrt2}2\right)-\\ & \frac14\sqrt{2+\sqrt2}\ln\left(\sqrt{\tan x}-\sqrt[4]{\tan x}\sqrt{2-\sqrt2}+1\right)-\\ & \frac12\sqrt{2-\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4-2\sqrt2}-\frac{2-\sqrt2}2\right)+\\ & \frac14\sqrt{2+\sqrt2}\ln\left(\sqrt{\tan x}+\sqrt[4]{\tan x}\sqrt{2-\sqrt2}+1\right)-\\ & \frac12\sqrt{2-\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4-2\sqrt2}+\frac{2-\sqrt2}2\right)-\\ & \frac14\sqrt{2-\sqrt2}\ln\left(\sqrt{\tan x}+\sqrt[4]{\tan x}\sqrt{2+\sqrt2}+1\right)+\\ & \frac12\sqrt{2+\sqrt2}\tan^{-1}\left(\sqrt[4]{\tan x}\sqrt{4+2\sqrt2}+\frac{2+\sqrt2}2\right)+C\end{align}$$ All in all, combining complex conjugates before integrating seemed to work well in this case. It took much more time to typeset the solution than to solve the integral by hand. It agrees with the Mathematica solution already posted.
user5713492
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[Too long for a comment]
Mathematica gives this answer.
$\displaystyle\int \sqrt[4]{\tan (x)} \, dx=\\\cos \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}-\cos \left(\frac{\pi }{8}\right)\right)\right)+\\\cos \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\csc \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}+\cos \left(\frac{\pi }{8}\right)\right)\right)-\\\sin \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}-\sin \left(\frac{\pi }{8}\right)\right)\right)-\\\sin \left(\frac{\pi }{8}\right) \tan ^{-1}\left(\sec \left(\frac{\pi }{8}\right) \left(\sqrt[4]{\tan (x)}+\sin \left(\frac{\pi }{8}\right)\right)\right)+\\\frac{1}{2} \cos \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}+2 \sin \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)-\\\frac{1}{2} \cos \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}-2 \sin \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)-\\\frac{1}{2} \sin \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}+2 \cos \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)+\\\frac{1}{2} \sin \left(\frac{\pi }{8}\right) \log \left(\sqrt{\tan (x)}-2 \cos \left(\frac{\pi }{8}\right) \sqrt[4]{\tan (x)}+1\right)+C$
I wouldn’t want to have to have worked it out by hand, but it doesn’t look impossible.
Steve Kass
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