I have recently come across iterated integrals of the form $$\int_0^tdt_1(t-t_1)^{-1/2}\int_0^{t_1}dt_2 (t_1-t_2)^{-1/2}\cdots\int_{0}^{t_{m-1}}dt_m(t_{m-1}-t_m)^{-1/2}. \tag{1}$$ I am wondering if there is a closed formula for these expressions for general $m$. If $m=2$, then the expression above reduces to $$2\int_0^t dt_1 (t-t_1)^{-1/2} t_1^{1/2} = 2t\int_0^1d\tau(1-\tau)^{-1/2}\tau^{1/2} = \pi t.$$
But it's not clear to me how to generalize this to $m>2$.
Do the substitution $x_1=t-t_1,\ x_2=t_1-t_2,\ \dots,\ x_m=t_{m-1}-t_m$.
The integral becomes $D_m(1/2,\dots,1/2;t)$ where, for $a_1,\dots,a_m,x>0$, $$D_m(a_1,\dots,a_m;x)=\idotsint\limits_{\substack{x_1,\ \dots,\ x_m\geqslant 0\\x_1+\dots+x_m\leqslant x}}x_1^{a_1-1}\cdots x_m^{a_m-1}\,dx_1\cdots\,dx_m.$$
Known as Dirichlet's beta integral, this can be evaluated as follows. Clearly $$D_m(a_1,\dots,a_m;x)=x^{a_1+\dots+a_m}D_m(a_1,\dots,a_m;1)$$ (seen after substituting $x_k=xy_k$). Hence \begin{align*} \int_0^\infty e^{-x}D_m(a_1,\dots,a_m;x)\,dx &=\Gamma(a_1+\dots+a_m+1)D_m(a_1,\dots,a_m;1) \\&=\idotsint\limits_{x_1,\ \dots,\ x_m\geqslant 0\\\color{blue}{x\geqslant x_1+\dots+x_m}}x_1^{a_1-1}\cdots x_m^{a_m-1}\color{blue}{e^{-x}}\,dx_1\,\cdots\,dx_m\,\color{blue}{dx} \\&=\idotsint\limits_{x_1,\ \dots,\ x_m\geqslant 0}x_1^{a_1-1}\cdots x_m^{a_m-1}e^{-x_1-\dots-x_m}\,dx_1\,\cdots\,dx_m \\&=\Gamma(a_1)\cdots\Gamma(a_m) \\\implies D_m(a_1,\dots,a_m;1)&=\frac{\Gamma(a_1)\cdots\Gamma(a_m)}{\Gamma(a_1+\dots+a_m+1)} \end{align*} and the answer is $D_m(1/2,\dots,1/2;t)=(\pi t)^{m/2}/\Gamma(1+m/2)$.