$\int_{0}^{1} \int_{y}^{\sqrt{y}} \frac{\operatorname{sen}(x)}{2 x} d x d y$

Step 1 ：First, we need to understand the meaning of the double integral. It represents the volume under the surface defined by the function \(\frac{\operatorname{sen}(x)}{2x}\) over the region in the xy-plane bounded by \(0 \leq y \leq 1\) and \(y \leq x \leq \sqrt{y}\).

Step 2 ：Next, we need to calculate the inner integral first. The inner integral is \(\int_{y}^{\sqrt{y}} \frac{\operatorname{sen}(x)}{2x} dx\).

Step 3 ：However, this integral is not easy to solve directly. We can use the method of integration by parts, where we let \(u = \operatorname{sen}(x)\) and \(dv = \frac{1}{2x} dx\). Then we have \(du = \operatorname{cos}(x) dx\) and \(v = \frac{1}{2} \ln|x|\).

Step 4 ：Applying the integration by parts formula, we get \(\int_{y}^{\sqrt{y}} \frac{\operatorname{sen}(x)}{2x} dx = \left[\frac{1}{2} \operatorname{sen}(x) \ln|x|\right]_{y}^{\sqrt{y}} - \int_{y}^{\sqrt{y}} \frac{1}{2} \operatorname{cos}(x) \ln|x| dx\).

Step 5 ：Substitute the limits of integration into the equation, we get \(\frac{1}{2} \operatorname{sen}(\sqrt{y}) \ln|\sqrt{y}| - \frac{1}{2} \operatorname{sen}(y) \ln|y| - \int_{y}^{\sqrt{y}} \frac{1}{2} \operatorname{cos}(x) \ln|x| dx\).

Step 6 ：Then, we substitute this result into the outer integral, we get \(\int_{0}^{1} \left(\frac{1}{2} \operatorname{sen}(\sqrt{y}) \ln|\sqrt{y}| - \frac{1}{2} \operatorname{sen}(y) \ln|y| - \int_{y}^{\sqrt{y}} \frac{1}{2} \operatorname{cos}(x) \ln|x| dx\right) dy\).

Step 7 ：However, this integral is still not easy to solve directly. We can use numerical methods to approximate the value of this integral.