Integral of $\int_0^1 \frac{dx}{(1-x^5)^{1/5}}$

Question

I've attempted to integrate $I = \int_0^1\frac {dx}{(1-x^5)^{1/5}}$ using residues. I've convinced myself that this integral converges, and am under the impression that $f(z) = \frac{1}{(1-z^5)^{1/5}}$ has branch points at the five fifth roots of unity. I've taken the rays from these points out to infinity as my branch cuts, and attempted to calculate the integral along a contour $0 \to 1 \to e^{2\pi i/5} \to 0$, where I've made little quarter circles around the singularities. It seems to me (from a calculation) that the little circles around the singularities have zero contribution in the limit $r\to 0$. It also seems that the integral $e^{2\pi i/5}\to 0$ is just $-e^{2\pi i/5} I$, and finally the arc between the singularities is proving a little tricky to me (initially I tried to ignore convergence issues and use a binomial series but this may well be where I went wrong...). Any tips are appreciated!

Answer 1

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\B}{\mathscr{B} }$I can't see what you're seeing because you haven't shown exactly what you've done. Let me instead offer some ideas for the computation of this. The second item - dogbone contour integration - is very useful and I hope will interest you.

I definitely don't like to see $x^5$ there. Let's take $x\mapsto x^{1/5}$ to instead calculate: $$\frac{1}{5}\int_0^1(1-x)^{-1/5}x^{-4/5}\d x$$

I like this because I immediately recognise it as a value of the Beta function: $$\frac{1}{5}\B(4/5,1/5)=\frac{1}{5}\frac{\Gamma(1/5)\Gamma(4/5)}{\Gamma(1)}=\frac{\pi}{5}\csc\frac{\pi}{5}$$

Can we do this with complex analysis? Yes. See, another reason that I want to take $x\mapsto x^{1/5}$ in the integrand is so that I can get two powers. Now complex integration with one fractional power is sometimes a pain due to those pesky branch cuts, but when you have two you can combine them so the cuts "cancel" out, leading to the notion of "dogbone contour".

So let $D$ be the contour that winds clockwise $i\epsilon\to1+i\epsilon\to1-i\epsilon\to-i\epsilon\to i\epsilon$ where some of those paths are semicircles of radius $\epsilon$ and $\epsilon>0$. It looks like a race track, or a "dogbone".

Define $(1-z)^{-1/5}$ via the the logarithm with argument $-\pi\le\arg<\pi$ and $z^{-4/5}$ via the logarithm with argument $0\le\arg<2\pi$. There are branch cuts on $[1,\infty)$ and $[0,\infty)$ respectively, but we'll see that they "cancel" to a branch cut merely on $[0,1]$.

In particular: $$f:z\mapsto\frac{1}{5}(1-z)^{-1/5}z^{-4/5}$$Is now holomorphic outside of $D$. To see why, I invite you to check that for any real $x>1$, $\lim_{z\to x}f(z)$ exists and equals $f(x)$: the discontinuity due to argument-jumping is no longer present and there are no challenges to $f$ being holomorphic.

Thus the integral around $D$ is $2\pi i$ multiplied by the residue at infinity. What is that? Well, note $f$ vanishes as $|z|\to\infty$ and so it follows the residue is equal to $\lim_{|z|\to\infty}-zf(z)$, which exists by some formal analysis and thus can be evaluated along any path. In particular, take the path as $z\to\infty$ along the positive real axis. Then when $z>1$ we get $f(z)=\frac{1}{5}e^{(-\pi i)(-1/5)}(x-1)^{-1/5}x^{-4/5}$ and then the limit of $-xf(x)$ is clearly: $$-\frac{1}{5}e^{\pi i/5}$$

Alright. What happens if we squeeze $\epsilon\to0^+$? The integrals around the semicircles at $0,1$ vanish (not hard to see) and the integral along $i\epsilon\to1+i\epsilon$ tends to: $$\frac{1}{5}\int_0^1(1-x)^{-1/5}x^{-4/5}\d x=I$$Whereas the integral along $1-i\epsilon\to-i\epsilon$ tends to (mind the argument jump on $(0,1)$! That’s partially why this works at all, otherwise we’d just get $I-I=0$ which is trivial): $$\frac{1}{5}\int_0^1(1-x)^{-1/5}x^{-4/5}(e^{-2\pi i\cdot 4/5})\d x=e^{-8\pi i/5}I$$Thus: $$(1-e^{-8\pi i/5})I=(2\pi i)\cdot\underset{\operatorname{Res}_\infty}{\underbrace{\left(-\frac{1}{5}e^{\pi i/5}\right)}}$$Concluding: $$I=\frac{\pi}{5}\csc\frac{\pi}{5}$$After a little algebra. Specifically: $$I=\frac{\pi}{5}\frac{-e^{\pi i/5}}{(1-e^{-8\pi i/5})/2i}=\frac{\pi}{5}\frac{-e^{5\pi i/5}}{(e^{4\pi i/5}-e^{-4\pi i/5})/2i}=\frac{\pi}{5}\frac{1}{\sin(4\pi/5)}=\frac{\pi}{5}\csc\frac{\pi}{5}$$

Answer 2

I have tried to use the binomial for the parenthesis $$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$$ and then replaced the 1 and used the generalized form because k is not integer $$(1+x)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^k$$ where $$\binom{r}{k} = \frac{r(r-1)(r-2)...(r-k+1)}{k!}$$ $$(1-x^5)^{1/5} = \sum_{k=0}^{\infty} \binom{1/5}{k} (-x^5)^k$$ integrating before evaluation would result in $$\int (1-x^5)^{1/5} dz = \sum_{k=0}^{\infty} \binom{1/5}{k} \frac{(-x^5)^{k+1}}{5(k+1)} + C$$ and with evaluation: $$\int_{0}^{1} (1-x^5)^{1/5} dx = \left[ \sum_{k=0}^{\infty} \binom{1/5}{k} \frac{(-x^5)^{k+1}}{5(k+1)} \right]_{0}^{1} = \left[ \sum_{k=0}^{\infty} \binom{1/5}{k} \frac{(-1)^{k+1}}{5(k+1)} \right] - \left[ \sum_{k=0}^{\infty} \binom{1/5}{k} \frac{(0)^{k+1}}{5(k+1)} \right]$$ and that results in the final: $$\int_{0}^{1} (1-x^5)^{1/5} dx = \sum_{k=0}^{\infty} \binom{1/5}{k} \frac{(-1)^{k+1}}{5(k+1)}$$ which converges from 0 to 1 Let me know if i have any mistakes or the answer is not short enough

Answer 3

I have not found the solution but I think I've managed to make an integral look easier I used a substitution and plugged it in with a handful of easy algebra enter image description here
Hope this answer helps somehow... P.S. Sorry, this is my first try making a post on this site so I dont know how you guys all write these beautiful math formulas in here so i just added a picture with formulas i painfully typed in Word.