If this area is finite, what is the value of area of this integral?

\( \int \limits_1^{\infty}\frac{\ln x}{x^2}\, dx=\lim_{t\to\infty}\int \limits_1^t \frac{\ln x}{x^2}\\\\ \int \frac{\ln x}{x^2}\, dx=(*)\\ u=\ln x, du=\frac{1}{x} \\ dv=\frac{1}{x^2}, v=-\frac{1}{x}\\ (*)=\ln x\cdot(-\frac{1}{x})-\int (\frac{1}{x}\cdot(-\frac{1}{x}))\, dx=\\ -\frac{\ln x}{x}+\int\frac{1}{x^2}\, dx=\\ -\frac{\ln x}{x}-\frac{1}{x}+C\\\\ \lim_{t\to\infty}\int \limits_1^t \frac{\ln x}{x^2}=\lim_{t\to \infty}\left[-\frac{\ln x}{x}-\frac{1}{x} \right]_1^t= \)

\( \lim_{t\to \infty}\left(-\frac{\ln t}{t}-\frac{1}{t}-\left(-\frac{\ln 1}{1}-\frac{1}{1}\right)\right)=\\ \lim_{t\to \infty}\left(-\frac{(\ln t)â}{tâ}\right)-0-(-1)=\\ \lim_{t\to \infty}\left(-\frac{\frac{1}{t}}{1}\right)+1=\\ \lim_{t\to \infty}\left(-\frac{1}{t}\right)+1=\\ 0+1=\\ 1 \)