To solve the problem, we need to evaluate the integral \( I = \int_{1}^{e} f''(x) \ln x \, dx \) given the conditions \( f'(e) = f(e) = f(1) = 1 \) and \( \int_{1}^{e} \frac{f(x)}{x^2} \, dx = \frac{1}{2} \).
### Step-by-Step Solution:
1. **Integration by Parts**: We will use integration by parts on the integral \( I = \int_{1}^{e} f''(x) \ln x \, dx \). We set:
- \( u = \ln x \) (which gives \( du = \frac{1}{x} dx \))
- \( dv = f''(x) dx \) (which gives \( v = f'(x) \))
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[
I = \left[ \ln x \cdot f'(x) \right]_{1}^{e} - \int_{1}^{e} f'(x) \cdot \frac{1}{x} \, dx
\]
2. **Evaluate the Boundary Terms**: Now we evaluate the boundary terms:
- At \( x = e \): \( \ln e = 1 \) and \( f'(e) = 1 \), so \( \ln e \cdot f'(e) = 1 \cdot 1 = 1 \).
- At \( x = 1 \): \( \ln 1 = 0 \) and \( f'(1) \) is unknown, but it will be multiplied by 0, so this term contributes 0.
Thus, the boundary term evaluates to:
\[
\left[ \ln x \cdot f'(x) \right]_{1}^{e} = 1 - 0 = 1
\]
3. **Substituting Back into the Integral**: Now we substitute back into the integral:
\[
I = 1 - \int_{1}^{e} \frac{f'(x)}{x} \, dx
\]
4. **Evaluate the Remaining Integral**: We need to evaluate \( \int_{1}^{e} \frac{f'(x)}{x} \, dx \). We can use the substitution \( f(x) = x^2 g(x) \) (where \( g(x) = \frac{f(x)}{x^2} \)). The given condition states:
\[
\int_{1}^{e} \frac{f(x)}{x^2} \, dx = \frac{1}{2}
\]
This implies:
\[
\int_{1}^{e} g(x) \, dx = \frac{1}{2}
\]
5. **Relate \( f'(x) \) to \( g(x) \)**: We know that:
\[
f'(x) = 2x g(x) + x^2 g'(x)
\]
Thus,
\[
\int_{1}^{e} \frac{f'(x)}{x} \, dx = \int_{1}^{e} \left( 2g(x) + x g'(x) \right) \, dx
\]
6. **Evaluate the Integral**:
- The first part gives \( 2 \int_{1}^{e} g(x) \, dx = 2 \cdot \frac{1}{2} = 1 \).
- The second part \( \int_{1}^{e} g'(x) \, dx = g(e) - g(1) \).
Since \( g(1) = \frac{f(1)}{1^2} = 1 \) and \( g(e) = \frac{f(e)}{e^2} = \frac{1}{e^2} \), we have:
\[
\int_{1}^{e} g'(x) \, dx = \frac{1}{e^2} - 1
\]
7. **Combine Results**: Therefore,
\[
\int_{1}^{e} \frac{f'(x)}{x} \, dx = 1 + \left( \frac{1}{e^2} - 1 \right) = \frac{1}{e^2}
\]
8. **Final Calculation**: Substitute this back into our expression for \( I \):
\[
I = 1 - \frac{1}{e^2}
\]
### Conclusion:
The value of the integral \( \int_{1}^{e} f''(x) \ln x \, dx \) is:
\[
\boxed{1 - \frac{1}{e^2}}
\]
