Find a number $\alpha > 1$ such that the following holds 
We want to find a real number $\alpha > 1$ such that 
$$ \int\limits_{0}^{\infty} \frac{ dx }{(1+x^{\alpha})^{\alpha}} = 1
 $$

What I got so far
First, I used Integration by parts to obtain 
$$ \int\limits_{0}^{\infty} \frac{ dx }{(1+x^{\alpha})^{\alpha}} = \underbrace{\frac{x}{(1+x^{\alpha})^{\alpha}} \bigg|_0^{\infty}}_{=0} + \int\limits_0^{\infty} \frac{ \alpha^2 x^{\alpha} }{(1+x^{\alpha})^{\alpha+1}} dx $$
Now, notice
$$ \int\limits_0^{\infty} \frac{ \alpha^2 (x^{\alpha}  +1 -1)}{(1+x^{\alpha})^{\alpha+1}} dx = \int\limits_0^{\infty} \frac{ \alpha^2  }{(1+x^{\alpha})^{\alpha}} dx - \int\limits_0^{\infty} \frac{ \alpha^2  }{(1+x^{\alpha})^{\alpha+1}} dx  $$
Therefore, for our original integral to be $1$, we must have 
$$ \int\limits_0^{\infty} \frac{ \alpha^2  }{(1+x^{\alpha})^{\alpha}} dx =1+ \int\limits_0^{\infty} \frac{ \alpha^2  }{(1+x^{\alpha})^{\alpha+1}} dx $$
But, notice
$$ 1 = - \int\limits_0^{\infty} \frac{ \alpha x^{\alpha - 1} dx }{(1+ x^{\alpha} )^{\alpha}} $$
Thus, the above expression is equivalent to
$$ \int\limits_0^{\infty} \frac{\alpha^2+\alpha x^{\alpha-1}}  {(1+x^{\alpha})^{\alpha}} = \int\limits_0^{\infty} \frac{ \alpha^2  }{(1+x^{\alpha})^{\alpha+1}} dx $$
Therefore,
$$ (1+x^{\alpha})^{\alpha}(\alpha^2 + \alpha x^{\alpha-1} ) = \alpha^2 $$
holds for all $x>0$. In particular, if $x=1$, we have 
$$ 2(\alpha^2 + \alpha) = \alpha^2 $$
but this equation has no solution  $\alpha>1$. What am I doing wrong?
 A: I try some numbers in Wolframalpha and came up with the conjecture (I'm very sure is true).
\begin{align}
\int^\infty_0 \frac{dx}{(1+x^\alpha)^\alpha} = \frac{\Gamma(\alpha-\frac{1}{\alpha})\Gamma(1+\frac{1}{\alpha})}{\Gamma(\alpha)}.
\end{align}
If you set 
\begin{align}
\frac{\Gamma(\alpha-\frac{1}{\alpha})\Gamma(1+\frac{1}{\alpha})}{\Gamma(\alpha)}=1
\end{align}
we see that there is a solution on the interval $(1, 2)$. Solving for $\alpha$ yields
\begin{align}
\alpha = \frac{1+\sqrt{5}}{2}.
\end{align}
Edit: Observe
\begin{align}
\int^\infty_0 \frac{dx}{(1+x^\alpha)^\alpha} =&\ \alpha \int^\infty_0 \frac{\alpha x^\alpha}{(1+x^\alpha)^{\alpha+1}}\ dx
 =\ \alpha \int^\infty_0 \frac{t^{1/\alpha}}{(1+t)^{1+\alpha}}\ dt\\
 =&\ \alpha B(1+1/\alpha, \alpha-1/\alpha) =\ \alpha\frac{\Gamma(\alpha-\frac{1}{\alpha})\Gamma(1+\frac{1}{\alpha})}{\Gamma(\alpha+1)}\\
=&\ \frac{\Gamma(\alpha-\frac{1}{\alpha})\Gamma(1+\frac{1}{\alpha})}{\Gamma(\alpha)}.
\end{align}
A: Your equation
$$ 1 = - \int\limits_0^{\infty} \frac{ \alpha x^{\alpha - 1} dx }{(1+ x^{\alpha} )^{\alpha}} $$
can't be correct. Note that the integral of a positive function is again positive. Instead,
$$\int\limits_0^{\infty} \frac{ \alpha x^{\alpha - 1} dx }{(1+ x^{\alpha} )^{\alpha}} = \left.\frac{(x^\alpha + 1)^{1 - \alpha}}{1 - \alpha} \right|_0^\infty = \frac{1}{\alpha - 1}$$
is correct.
A: $\alpha=\frac{1+\sqrt 5}{2}$ is the solution.
Edit because of the downvote:
$\int\dfrac{dx}{(1+x^\alpha)^\alpha}=\int \dfrac{dx}{(1+x^\frac{1}{\alpha-1})^\alpha}=x(x^{\frac{1}{\alpha-1}}+1)^{1-\alpha}=x(x^\alpha+1)^{1-\alpha}$
So $\int^\infty_0\dfrac{dx}{(1+x^\alpha)^\alpha}=\lim_{x\rightarrow\infty} x(x^\alpha+1)^{1-\alpha}=\lim_{x\rightarrow\infty} \frac{x}{(x^\alpha+1)^\frac{1}{\alpha}}=1$
Edit: $\alpha=\frac{1}{\alpha-1}$ as $\alpha$ is the golden ratio
A: If $\alpha = \phi\textrm{( the golden ratio )}, \textrm{ then }\alpha^2-\alpha -1=0,$
$$
\begin{aligned}
\int_0^{\infty} \frac{1}{\left(1+x^\alpha\right)^\alpha} d x & \stackrel{x\mapsto\frac{1}{x}}{=}  \int_{\infty}^0 \frac{1}{\left(1+\frac{1}{x^\alpha}\right)^\alpha} \frac{d x}{-x^2} \\
& =\int_0^{\infty} \frac{x^{\alpha^2-2}}{\left(x^\alpha+1\right)^\alpha} d x \\
& =\int_0^{\infty} \frac{x^{\alpha-1}}{\left(x^\alpha+1\right)^\alpha} d \alpha \\
& =\frac{1}{\alpha} \int_0^{\infty} \frac{d\left(x^\alpha+1\right)}{\left(x^\alpha+1\right)^\alpha} \\
& =\frac{1}{\alpha}\left[\frac{\left(x^\alpha+1\right)^{\alpha+1}}{-\alpha+1}\right]_0^{\infty} \\
& =\frac{1}{\alpha(1-\alpha)} {(-1)} \\
& =1
\end{aligned}
$$
