It is well-known that for a standard Brownian motion $(B_t)_{t\geq 0}$ the first hitting time of the level $1$ $$T_1 := \inf\{t> 0 : B_t = 1 \}$$ has standard Levy distribution, this means that $T_1$ has the density $$\sqrt{\frac{1}{2\pi t^3}} \exp\left(-\frac{1}{2t}\right)$$
It is also known that $\Bbb E[T_1^n] = \infty $. But is there a closed form expression for $$\Bbb E[(t-T_1)^n 1_{\{T_1 \leq t\}}] = \int_0^t (t-u)^{n} \sqrt{\frac{1}{2\pi u^3}} \exp\left(-\frac{1}{2u}\right) du$$ available?
Context: Im interested in the Laplace transform of the occupation time $\Gamma_t := \int_0^t 1_{(-\infty , 1]}(B_s) ds$ for $\lambda \leq 0$.
The distribution of the occupation time $\int_0^t 1_{(-\infty , 0]}(B_s) ds$ has the arcsin distribution by the Levys arcsin law. The laplace transform of the standard arcsin distribution is the function $_1F_1(1/2,1, \lambda).$ Thus we have by $\Gamma_t = T_1 \wedge t + \int_{T_1 \wedge t}^t 1_{(-\infty , 1]}(B_s) ds$ and the Markov property that the Laplace transform of $\Gamma_t $ is given by $$e^{\lambda t}\Bbb P (T_1 > t) + \int_{0}^t e^{\lambda u} \sqrt{\frac{1}{2\pi u^3}} \exp\left(-\frac{1}{2u}\right) {}_1F_1(1/2,1, \lambda (t-u)) du$$
Note that $_1F_1(1/2,1, z) = \sum_{n=0}^\infty \frac{(1/2)^{(n)} z^n}{n!}$.
For $n \in \mathbb{N}^+$ and $t>0$, Mathematica gives:
$$t^n \left(\, _1F_1\left(-n;\frac{1}{2};-\frac{1}{2 t}\right)-\frac{\sqrt{2} \Gamma (n+1) \, _1F_1\left(\frac{1}{2}-n;\frac{3}{2};-\frac{1}{2 t}\right)}{\sqrt{t} \Gamma \left(n+\frac{1}{2}\right)}\right)$$
where $F$ is the confluent hypergeometric function of the first kind.
For $n = 2$, here is your function of $t$:
Hope this helps!