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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?

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    Once you have one representation for $f^{-1}(x)$, one example of such an integral is $\int_0^1f^{-1}(x)dt$. I'm pretty sure that inverse doesn't exist in terms of elementary functions though. Also note that you will need a restricted domain for the inverse, since $x\cot x$ is not injective. – 2017-01-06
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    Yes. I should have specified what domain I'm interested in. I'll edit and put in some details. – 2017-01-06
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    We have $\displaystyle x\cot x=1-2\sum_1^\infty\zeta(2n)(\frac{x}{\pi})^{2n}$ and is not an elementary function. – 2017-01-06
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    @user3798897 - I made some changes to the question. – 2017-01-08
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    @MyGlasses - I am not restricting myself to elementary functions... – 2017-01-08

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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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