Question

\rm\color{red}{\large \int_0^\infty (x^(\frac\pi5-1))/(1+x^(2\pi)) \mathrm dx} \\

asked Apr 4, 2023 82.2k views

1 Answer

Chnoch · Answer 1 · 2023-04-08T21:18:27+0000

Substitute
$y=x^(\pi/5)$ and
$dy=\frac\pi5 x^(\pi/5 - 1) \, dx$ .

$\displaystyle \int_0^\infty (x^(\frac\pi5-1))/(1+x^(2\pi)) \, dx = \frac5\pi \int_0^\infty (dy)/(1+y^(10))$

Now substitute
$y=\left(\frac1z-1\right)^(1/10)$ and
$dy=-\frac1{10z^2} \left(\frac1z-1\right)^(-9/10) \, dz$ to get a beta integral.

$\displaystyle \int_0^\infty (x^(\pi/5-1))/(1+x^(2\pi)) \, dx = \frac1{2\pi} \int_0^1 z^(-1/10) (1-z)^(-9/10) \, dz \\\\ ~~~~~~~~~~~~~~~~~~~~~ = \boxed{\frac1{2\pi} \, \mathrm{B}\left(\frac9{10}, \frac1{10}\right)}$

We can do better:

Recall that

$\mathrm{B}(a,b) = (\Gamma(a) \Gamma(b))/(\Gamma(a+b))$

as well as the reflection formula for the gamma function,

$\Gamma(z) \Gamma(1-z) = (\pi)/(\sin(\pi z))$

It follows that

$\displaystyle \int_0^\infty (x^(\pi/5-1))/(1+x^(2\pi)) \, dx = \frac1{2\pi} \cdot \frac{\pi}{\sin\left(\frac\pi{10}\right) \Gamma(1)} = \boxed{\frac12 \csc\left(\frac\pi{10}\right)}$

Even better:

To find an exact value for this result, recall

$\sin^2(\theta) = \frac{1-\cos(2\theta)}2$

and

$\cos(5\theta) = 5\cos(\theta) - 20 \cos^3(\theta) + 16 \cos^5(\theta)$

Then

$\theta=\pi \implies -1 = 5\cos\left(\frac\pi5\right) - 20 \cos^3\left(\frac\pi5\right) + 16 \cos^5\left(\frac\pi5\right)$

Let
$t=\cos\left(\frac\pi5\right)$ . Solve for
in the quintic equation.

$16t^5 - 20t^3 + 5t + 1 = 0 \\\\ 16 (t^5 - t^3) - 4(t^3 - t) + t+1 = 0 \\\\ (t+1) \left(16t^3 (t-1) + 4t(t-1) + 1\right) = 0 \\\\ (t+1) \left(16t^4 - 16t^3 - 4t^2 + 4t + 1\right) = 0 \\\\ (t + 1) \left(4t^2 - 2t - 1\right)^2 = 0$

Clearly
$t\\eq-1$ , so we're left with

$4t^2 - 2t - 1 = 0 \implies t = \frac{1 \pm \sqrt5}4$

and
t>0 , so we take the positive root.

Now

$\sin^2\left(\frac\pi{10}\right) = \frac{1 - \frac{1+\sqrt5}4}2 = \frac{3-\sqrt5}8 \\\\ ~~~~ \implies \sin\left(\frac\pi{10}\right) = \frac12 \sqrt{\frac{3-\sqrt5}2}$

Denest the radical. Suppose there are rational
a,b,c such that

$\sqrt{\frac{3-\sqrt5}2} = a + b\sqrt c$

Squaring both sides gives

$\frac{3-\sqrt5}2 = a^2 + b^2c + 2ab\sqrt c$

Let
c=5 . Solve for
a,b .

$\begin{cases} 2a^2+10b^2 = 3 \\ 4ab = 1 \end{cases} \implies b = -\frac1{4a}$

$a^2 + \frac5{16a^2} = \frac32 \implies 16a^4 - 24a^2 + 5 = 0 \\\\ ~~~~ \implies (4a^2 - 5) (4a^2 - 1) = 0 \\\\ ~~~~ \implies a^2 = \frac54 \text{ or } a^2 = \frac14$

The first case leads to irrational
, so we must have one of
$a=\pm\frac12$ and
$b=\mp\frac12$ . The value of
$\sqrt{\frac{3-\sqrt5}2}$ must be positive, which is consistent with
$a=-\frac12$ and
$b=\frac12$ .