evaluate $\\int_{1}^{3}\\int_{1}^{2}\\frac{\\ln y}{xy}dydx$.

Explanation:

Step1: Separate the integrals

Since the integrand (\frac{\ln y}{xy}) can be written as (\frac{1}{x}\cdot\frac{\ln y}{y}), we can separate the double - integral (\int_{1}^{3}\int_{1}^{2}\frac{\ln y}{xy}dydx) into the product of two single - integrals. By the property of double - integrals (\int_{a}^{b}\int_{c}^{d}f(x)g(y)dydx=\int_{a}^{b}f(x)dx\int_{c}^{d}g(y)dy), we have (\int_{1}^{3}\frac{1}{x}dx\int_{1}^{2}\frac{\ln y}{y}dy)

Step2: Evaluate (\int_{1}^{3}\frac{1}{x}dx)

Using the formula (\int\frac{1}{x}dx=\ln|x| + C), then (\int_{1}^{3}\frac{1}{x}dx=\left[\ln x\right]_{1}^{3}=\ln 3-\ln 1=\ln 3)

Step3: Evaluate (\int_{1}^{2}\frac{\ln y}{y}dy)

Let (u = \ln y), then (du=\frac{1}{y}dy). When (y = 1), (u=\ln 1 = 0); when (y = 2), (u=\ln 2) (\int_{1}^{2}\frac{\ln y}{y}dy=\int_{0}^{\ln 2}u du) Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), here (n = 1), so (\int_{0}^{\ln 2}u du=\left[\frac{u^{2}}{2}\right]_{0}^{\ln 2}=\frac{(\ln 2)^{2}}{2}-0=\frac{(\ln 2)^{2}}{2})

Step4: Multiply the results of the two integrals

(\int_{1}^{3}\frac{1}{x}dx\int_{1}^{2}\frac{\ln y}{y}dy=\ln 3\times\frac{(\ln 2)^{2}}{2}=\frac{(\ln 2)^{2}\ln 3}{2})

Answer:

(\frac{(\ln 2)^{2}\ln 3}{2})