Let $ f:R \rightarrow R $ be continuous and the integral $ \int_{0}^{\infty} f(x) dx $ convergent. Test integral for convergence. $$ \int_{\Gamma{\left( e^9 \right)}}^{\infty} f(\frac{2}{\pi}x\arctan x) dx $$
We know $ \Gamma{\left( 2 \right)} = 1 $ and $ \Gamma $ is monotously increasing from 2 to $ \infty $, so it has an inverse function in this range.
$$ \int_{\Gamma{\left( e^9 \right)}}^{\infty} f(\frac{2}{\pi}x\arctan x) dx = \int_{e^9}^{\infty} f(\frac{2}{\pi}\Gamma^{-1}{\left( y \right)}\arctan \Gamma^{-1}{\left( y \right)}) (\Gamma^{-1}{\left( y \right)})' dx $$
What can I do next? I don't know how to simplify function argument.
This is a homework assignment.
Somewhat involved hint: I don't think the Gamma-function has much to do with this. Let's let $g(x) = (2/\pi)x\arctan x.$ Suppose $0 < a < b.$ Think of $a$ as fixed and $b\to \infty.$ Then, letting $x=g^{-1}(y),$ we have
$$\int_a^b f(g(x))\, dx = \int_{g(a)}^{g(b)} f(y)\, (g^{-1})'(y)\, dy.$$
I'm thinking $g(x)$ looks very much like $x$ for large $x,$ hence $g'(x)$ should be close to $1$ for such $x.$ Therefore $(g^{-1})'(y)$ should be close to $1$ for large $y.$ Of course if we replaced $(g^{-1})'(y)$ with $1$ above, we would have a convergent integral as $b\to \infty.$ I would guess Dirichlet's test will be useful in clinching this. This is all hand-wavey, but I have a feeling ...