Equation involving an integral depending on two parameters I have some difficulty to find possible solutions of the following equation:
$$\int_0^\tau dx \frac{1}{x^\alpha+1}=\beta$$
where $\tau \gt 0,$ $\alpha\in \mathbb N$ ($\alpha=1,2,3,\dots$) and $\beta$ a given real valued constant.
Is it possible to find values of $\alpha$ and $\tau$ satisfiyng the equation? 
Thanks
 A: For every $\alpha$ in $[0,1]$, the function $g_\alpha:\tau\mapsto\int\limits_0^\tau\frac{\mathrm dx}{1+x^\alpha}$ is increasing from $g_\alpha(0)=0$ to $\lim\limits_{\tau\to+\infty}g_\alpha(\tau)=+\infty$. Hence, for each $\beta\geqslant0$, there exists a unique $\tau$ such that $g_\alpha(\tau)=\beta$.
If $\alpha\gt1$, the same result holds provided $\beta\lt\ell_\alpha$, where $\ell_\alpha=\lim\limits_{\tau\to+\infty}g_\alpha(\tau)$ is finite.
A: We start with:
$$\int_0^\tau {\frac{1}{1+x^\alpha} dx}=\beta$$
$$\int_0^\tau {\frac{x}{x+x^{\alpha+1}} dx}$$
Now consider the following substitution...  Let
$$u=x, \qquad dv=\frac{1}{x+x^{\alpha+1}}dx$$
$$du=dx, \qquad v=\int{\frac{1}{x+x^{\alpha+1}}}dx$$
We get
$$\int {\frac{x}{x+x^{\alpha+1}} dx} = x \int{\frac{1}{x+x^{\alpha+1}}}dx - \int {\left( \int{\frac{1}{x+x^{\alpha+1}} dx }\right) dx}$$
We can then use the shortcut substitution found in this question.  Noting
$$\int { \frac{1}{x + x^\gamma} dx}=\int {\left( \frac{x^{\gamma-1}+1}{x + x^\gamma } - \frac{x^{\gamma-1}}{x + x^\gamma } \right) dx}=f_\gamma(x)$$
$$=\frac{\gamma \log(x) - \log(x + x^\gamma)}{\gamma - 1} = g_\gamma(x)$$
...we can set $\gamma = \alpha + 1$ and our previous equation becomes
$$\int {\frac{x}{x+x^{\alpha+1}} dx} = x \int{\frac{1}{x+x^{\alpha+1}}}dx - \int {\left( \int{\frac{1}{x+x^{\alpha+1}} dx }\right) dx}$$
$$= x g_\gamma(x) - \int { \int{\frac{1}{x+x^{\gamma}} dx } dx}$$
We can now make use of the Fractional Calculus, or more exactly, the well known Cauchy formula for repeated integration to simplify the double integral on the righthand side.  The general formula is:
$$(J^n h)(x) = \frac{1}{(n-1)!} \int_0^x{(x-t)^{n-1} h(t) dt}$$
...where $(J^n h)(x)$ denotes $n$ repeated integrations on the integral of $h(x)$.  Plugging in $\frac{1}{t+t^{\gamma}}$ for $h(t)$, we get
$$\int_0^x {(x-t) \frac{1}{t+t^{\gamma}} dt}$$
I'll do more work, but I'm stopping for lunch for now...
