correct limits of integration I just want to clarify where did the $1/4$ come from?
Let $X$ denote the diameter of an armored electric
cable and $Y$ denote the diameter of the ceramic
mold that makes the cable. Both $X$ and $Y$ are scaled
so that they range between $0$ and $1$. Suppose that $X$
and $Y$ have the joint density
$$f(x, y) =\begin{cases}
\frac1y,&0<x<y<1\\\\
0,&\text{elsewhere}
\end{cases}$$
Solution:
$$\begin{align*}
&P\left(X+Y>\frac12\right)=1-P\left(X+Y<\frac12\right)=1-\int_0^{1/4}\int_x^{1/2-x}\frac1y dy\,dx\\
&=\left. 1-\int_0^{1/4}\left[\ln\left(\frac12-x\right)-\ln x\right]dx=1+\left[\left(\frac12-x\right)\ln\left(\frac12-x\right)-x\ln x\right]\right\vert_0^{1/4}\\
&=1+\frac14\ln\left(\frac14\right)=0.6534.
\end{align*}$$
 A: Since $0\le X,Y\le 1$, the region on which $X+Y<\frac12$ is the the triangle bounded by the axes and the line $x+y=\frac12$. To integrate over this region, you’d normally set up the integral like this:
$$P\left(X+Y<\frac12\right)=\int_0^{1/2}\int_0^{1/2-x}f(x,y)dy\,dx\;.$$
However, in this case you know that $f(x,y)=0$ when $y\le x$, so you can ignore any part of the triangle lying below the diagonal $y=x$. Thus, the region over which you need to integrate is actually the triangle bounded by $x=0$, $y=x$, and $x+y=\frac12$, which is shaped roughly like this: $\triangleright$. Everywhere else, $f(x,y)=0$. The righthand vertex of that triangle is at the point $\left\langle\frac14,\frac14\right\rangle$, so you need only run $x$ from $0$ to $1/4$.
A: As pointed out by Brian M. Scott, the region over which we need to integrate is the triangle with corners $(0,0)$, $(\frac{1}{4},\frac{1}{4} )$, and $(0,\frac{1}{2})$.  I would like to point out a slightly different approach to the double integral that perhaps makes the integration easier.  
You integrated first with respect to $y$, and then with respect to $x$.  That has the disadvantage that you first get something that involves a couple of logarithms, which you then must integrate again.
Let us explore the alternative of integrating first with respect to $x$.  Then the first integration will be trivial, since $1/y$ can be treated as a constant.
The downside (look at the picture!) is that for $0 \le y \le \frac{1}{4}$, $x$ goes from $0$ to $y$, while for $\frac{1}{4}\le y\le \frac{1}{2}$, $x$ goes from $0$ to $\frac{1}{2}-y$.  So we will have to evaluate two integrals. We now do this, to show it is not hard.
First we evaluate
$$\int_{y=0}^{\frac{1}{4}}\left(\int_{x=0}^y\frac{dx}{y}\right)dy.$$
Very easily, the inner integral is $1$, so our integral is $\dfrac{1}{4}$.
Next we evaluate
$$\int_{y=\frac{1}{4}}^{\frac{1}{2}}\left(\int_{x=0}^{\frac{1}{2}-y}\frac{dx}{y}\right)dy.$$
The inner integral is $\dfrac{1}{2y}-1$, so our integral is $(1/2)(\ln(1/2)-\ln(1/4)) -(1/2-1/4)$.
This simplifies to $(1/2)\ln 2-1/4$.
Add the two parts. We get $(1/2)\ln 2$.  Finally, as in your calculation, the answer to the original question is $1-(1/2)\ln 2$.
Remark: That was easy, but in fact one can do better, by making the change of variable $w=x+y$. 
