I'm interested in determining a closed form expression of $f^{-1}(x)$ where $$ f(x) = \frac{\pi}{2} \sqrt{1-4x} \cot \left[ \frac{\pi}{2} \sqrt{1-4x} \,\right] ~. $$ Note that this function is one-to-one for $x>-\frac{3}{4}$. In this range, the function can be inverted. I'm hoping to find an integral representation of the inverse of the form
$f^{-1}(x) = \int_0^1 g(x,t) dt$
for some function $g(x,t)$. I honestly don't even know where to start or if it is even possible (not a HW question).
Any ideas?
I think this question very hard to solve, well, let $$g(x)=x\cot x$$ and $$h(x)=\frac{\pi}{2}\sqrt{1-4x}$$ so $$ f(x)=\frac{\pi}{2}\sqrt{1-4x}\cot\left[\frac{\pi}{2}\sqrt{1-4x}\,\right] =goh(x) $$ What's $f^{-1}=h^{-1}o g^{-1}$. we know $\displaystyle h^{-1}(x)=\frac{1}{4}-\Big(\frac{x}{\pi}\Big)^2$ for $\displaystyle x<\frac14$, so the problem is finding inverse of $g^{-1}$.
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