Question

Integration 0 to 1 dx/(1-x^6)^1/6
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Answer 1

Answer:

$\frac{\pi}{3}$

Step-by-step explanation:

Partially decomposing the integrand to fractions allows one to derive the integral in a simpler way. To prevent the introduction of any complex values, one can divide the integrand into quadratic parts rather than linear ones.

Mathematics heavily relies on functions. It is described as a unique correlation between a group of input and output values in which each input value corresponds to a specific output value. We are aware of two different categories of Euler integral functions. A gamma function and a beta function, respectively. The kind of function determines its domain, range, or codomain. We will talk about the definition, calculations, characteristics, and illustrations of beta functions on this page.

If one don't want to use complex analysis, one can evaluate the integral as a beta function.

For any $\alpha > -1, \beta > \alpha+1$. Let $y=x^\beta$  and $\gamma=\frac{\alpha+1}{\beta}$ . It is clear $0 < \gamma < 1$ and

$\int_0^{\infty} \frac{x^\alpha}{1+x^\beta} d x=\frac{1}{\beta} \int_0^{\infty} \frac{y^{\gamma-1}}{1+y} d y=\frac{1}{\beta} \int_0^{\infty}\left(\frac{y}{1+y}\right)^{\gamma-1}\left(\frac{1}{1+y}\right)^{2-\gamma} d y$

Let $z=\frac{y}{1+y}$ and notice $1-z=\frac{1}{1+y}$  and $d z=\frac{d y}{(1+y)^2}$, one can rewrite above integral into that for a beta function:

$\frac{1}{\beta} \int_0^1 z^{\gamma-1}(1-z)^{-\gamma} d z=\frac{1}{\beta} \frac{\Gamma(\gamma) \Gamma(1-\gamma)}{\Gamma(1)}=\frac{1}{\beta} \frac{\pi}{\sin \pi \gamma}=\frac{\pi}{\beta \sin \pi\left(\frac{\alpha+1}{\beta}\right)}$

For the question at hand, $\alpha=0, \beta=6$. The integral we want becomes

$\frac{\pi}{6 \sin \left(\frac{\pi}{6}\right)}=\frac{\pi}{3}$.

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