3

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

J. M. ain't a mathematician
  • 75,051
Riccardo.Alestra
  • 10,546
  • It is possible at least (i) for $\alpha=1$: you get $\log(\tau+1)=\beta$, (ii) for $\alpha=2$: you get $\tan^{-1}(\tau)=\beta$. Maybe you can extend this for composite $\alpha$... specially when $\alpha=2^n$... – draks ... Jul 19 '12 at 14:36
  • use Partial fraction expansion of $1/(x^4+1)$ and then integrate the parts like here – draks ... Jul 19 '12 at 14:43
  • W|A says that $\int \frac1{x^\alpha+1}dx=; _2F_1(1,\frac1n; 1+\frac1n, -x^n) +const$. I don't think that there is an analytical solution. Seems like all you can do is use did's proof of existence and numerics. Happy number crunching! – draks ... Jul 19 '12 at 18:28
  • @draks: I believe there is an analytical solution. Consider multiplying both the numerator and denominator by $x$ to get $\int {\frac{x}{x+x^{\alpha+1}}}$. Then use partial fractions to get this in terms of $\int {\frac{1}{x+x^{\alpha+1}}}$. Use the question posed here: http://math.stackexchange.com/questions/172124/ This is a shortcut to evaluate the integrals. You will also use the property of $\log{ab} = \log{a} + \log{b}$. I'm working on posting this as an answer... – Matt Groff Jul 20 '12 at 00:06
  • @MattGroff I've seen that question too and am looking forward to your answer... – draks ... Jul 20 '12 at 05:30

2 Answers2

4

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.

Did
  • 279,727
0

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...

Community
  • 1
Matt Groff
  • 6,117
  • @Riccardo.Alestra:Thanks for a very excellent question that was fun to try to answer. – Matt Groff Jul 20 '12 at 19:52
  • @draks: Let me know if you'd like to see more. I can always post as a seperate question,answer,etc... – Matt Groff Jul 20 '12 at 19:53
  • You are making circles: in the last displayed integral, the $x$ part of the parenthesis $(x-t)$ is $xg_\gamma(x)$ and the $t$ part is the integral $I$ you are trying to compute. And indeed $I=xg_\gamma(x)-(xg_\gamma(x)-I)$... Sorry about that. – Did Jul 20 '12 at 22:51