Just a Double Integral

Graph the region described in the iterated integral and recognize the equivalent region described by $0\leq y \leq x$ , $0 \leq x \leq 1$ . We use this to greatly simplify the problem $\begin{aligned} \int_0^1\int_y^1\frac{\sin(x)}{x}\,dx\,dy = & \int_0^x\int_0^1 \frac{\sin(x)}{x}\,dx\,dy \\ = & \int_0^1\int_0^x\frac{\sin(x)}{x}\,dy\,dx \\ = & \int_0^1 \sin(x)\,dx \\ = & 1-\cos(1). \end{aligned}$ So we have $a^b=1^1=\boxed{1}$ .

$\begin{aligned} I & = \int_0^1 \int_y^1 \frac {\sin x}x dx \ dy \\ & = \int_0^1 \left(\int_0^1 \frac {\sin x}x dx - \int_0^y \frac {\sin x}x dx \right) dy & \small \color{#3D99F6} \text{Since sine integral } \text{Si }(z) = \int_0^z \frac {\sin t}t dt \\ & = \int_0^1 \left(\text{Si }(1) - {\color{#3D99F6}\text{Si }(y)} \right) dy & \small \color{#3D99F6} \text{Using integration by parts } \\ & = y\text{Si }(1) \ \bigg|_0^1 - \color{#3D99F6} y\text{Si }(y) \ \bigg|_0^1 + \int_0^1 x \cdot \frac {\sin x}x dx \\ & = \text{Si }(1) - 0 - \text{Si }(1) + 0 - \cos y \ \bigg|_0^1 \\ & = 1 - \cos 1 \end{aligned}$

Therefore, $a^b = 1^1 = \boxed{1}$ .