Suppose $Y = \sqrt{2T}\cos(U)$, $ 0 \le u \le \pi $, and $ 0 \le \cos^{-1}(\frac{y}{\sqrt {2t}}) \le \pi ) $, so $ -1 \le \frac{y}{\sqrt{2t}} \le 1 $, with all $ \mathbb{R}$. The iterated integral is $$ G(y)= P(Y \le y) =P(\sqrt{2T}\cos(U) \le y)=\int_{0}^{y}\int_{0}^{\cos^{-1}(\frac{y}{\sqrt {2t}})} \frac{e^{-t}}{\pi } \mathrm{d}u \mathrm{d}t$$
My Mathematica code is
Integrate[((E^-t)/(\[Pi])), {u, 0, ArcCos[y/((2 t)^(1/2))]},
Assumptions ->
0 <= u <= \[Pi] && -1 <= ArcCos[y/((2 t)^(1/2))] <= 1 &&
Element[t, Reals] > 0]
Integrate[%, {t, 0, y},
Assumptions -> -1 <= ArcCos[y/((2 t)^(1/2))] <= 1 &&
Element[t, Reals] > 0]
but then Mathematica never seems to finish evaluating the second one, and so I have to Abort it. Is there any way to find this as a function of $y$? $G(y)$ is a CDF of $Y$.
Plot[ArcCos[1/Sqrt[2 t]], {t, 0, 1}]will immediately show that the inverse cosine is not even defined for sufficiently small values of $t$. The difficulty is that this problem--which appears to compute a probability by integration (assuming $T$ and $t$ are the same)--plays fast and loose with inverse trig functions and suffers accordingly. The solution is to re-solve the problem to generate an integral without the inverse cosine. – whuber Sep 6 '12 at 15:28