Recent content by TheFool

Actually, that integral isn't so simple. It cannot be stated as a finite combination of elementary functions.

I say that |x|<1/e since that is the radius of convergence of y=\sum_{m=0}^{\infty}\frac{(m-1)^{m-1}x^{m}}{m!}

Even though I showed I really am a fool for missing something so obvious, I decided to finish this. y=\sum_{m=0}^{\infty}\frac{(m-1)^{m-1}x^{m}}{m!} y'=\sum_{m=0}^{\infty}\frac{m^{m}x^{m}}{m!} Move on to the W function. -W(-x)=\sum_{m=1}^{\infty}\frac{m^{m-1}x^{m}}{m!} Differentiate...

Well, it would seem I don't belong posting in this forum. I shouldn't have missed that.

The Lagrange Inversion Theorem is used to find the inverse of a function. You didn't ask for the inverse in your original post :P

No wonder I'm confused by his substitutions.

Haruspex, could you use [tex] so we can better see what you're doing? It'll improve readability.

Another interesting thing of note, if f(x)=\sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!} then f'(x)=\sum_{n=0}^\infty \frac{n^{n}x^{n}}{n!} as long as |x|<1/e.

I doubt there is a closed form for it. Plus, it only converges if |x|<1/e. After looking at the graph, it's similar to the Lambert W function when |x|<1/e: -W(-x)=\sum_{m=1}^{\infty}{\frac{m^{m-1}x^{m}}{m!}} Subtracting 1 from both sides will make it approximately equal to your sum. However...

Is that substitution of y valid as it's an unknown function of x? What is u and v?

It seems unlikely that there will be a closed form solution for y since a and b can take on non-integer values as well. However, there may be a series representation of the logarithm of y. That can be integrated.

I said in a previous post: I'm talking about the non-trivial solution for y. It's almost a circle.

In fact, it almost looks like a circle. Putting in y=\sqrt{1-x^{2}} into the integral produces \int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{24}

That's how the problem was presented. Also, y is a function of x, not a constant. As such, it cannot be removed from the integral. I have implicitly plotted x^a-x^b=y^a-y^b for various values of a and b and I can see the two solutions for y. One is x, and the other fits the definition of...

In general, when you have something like f(x)^{g(x)} you can find the derivative like this, which should give you what you need to solve your specific problem where log is the natural logarithm: y=f(x)^{g(x)} \log{y}=g(x)\log{f(x)} Implicitly differentiate both sides...