Find the minimum value of $ \int_0^1 |e^{ \minus{} x} \minus{} a|dx\ ( \minus{} \infty < a < \infty)$.
To find the minimum value of \( \int_0^1 |e^{-x} - a| \, dx \) for \( -\infty < a < \infty \), we need to consider different cases for \( a \).

### Case 1: \( a \geq 1 \)
In this case, \( e^{-x} \leq 1 \leq a \) for all \( x \in [0, 1] \). Therefore, \( |e^{-x} - a| = a - e^{-x} \).

1. Compute the integral:
   \[
   \int_0^1 |e^{-x} - a| \, dx = \int_0^1 (a - e^{-x}) \, dx
   \]
2. Split the integral:
   \[
   \int_0^1 (a - e^{-x}) \, dx = \int_0^1 a \, dx - \int_0^1 e^{-x} \, dx
   \]
3. Evaluate each part:
   \[
   \int_0^1 a \, dx = a \int_0^1 1 \, dx = a \cdot 1 = a
   \]
   \[
   \int_0^1 e^{-x} \, dx = \left[ -e^{-x} \right]_0^1 = -e^{-1} + 1 = 1 - e^{-1}
   \]
4. Combine the results:
   \[
   \int_0^1 (a - e^{-x}) \, dx = a - (1 - e^{-1}) = a + e^{-1} - 1
   \]
5. Minimize the expression \( a + e^{-1} - 1 \):
   \[
   \text{This is minimized when } a = 1 \text{, giving the value } 1 + e^{-1} - 1 = e^{-1}
   \]

### Case 2: \( a \leq e^{-1} \)
In this case, \( e^{-x} \geq e^{-1} \geq a \) for all \( x \in [0, 1] \). Therefore, \( |e^{-x} - a| = e^{-x} - a \).

1. Compute the integral:
   \[
   \int_0^1 |e^{-x} - a| \, dx = \int_0^1 (e^{-x} - a) \, dx
   \]
2. Split the integral:
   \[
   \int_0^1 (e^{-x} - a) \, dx = \int_0^1 e^{-x} \, dx - \int_0^1 a \, dx
   \]
3. Evaluate each part:
   \[
   \int_0^1 e^{-x} \, dx = 1 - e^{-1}
   \]
   \[
   \int_0^1 a \, dx = a
   \]
4. Combine the results:
   \[
   \int_0^1 (e^{-x} - a) \, dx = (1 - e^{-1}) - a = 1 - e^{-1} - a
   \]
5. Minimize the expression \( 1 - e^{-1} - a \):
   \[
   \text{This is minimized when } a = e^{-1} \text{, giving the value } 1 - e^{-1} - e^{-1} = 1 - 2e^{-1}
   \]

### Case 3: \( e^{-1} < a < 1 \)
In this case, we need to split the integral at the point where \( e^{-x} = a \). Let \( a = e^{-b} \) for some \( b \in (0, 1) \).

1. Compute the integral:
   \[
   \int_0^1 |e^{-x} - a| \, dx = \int_0^b (e^{-x} - a) \, dx + \int_b^1 (a - e^{-x}) \, dx
   \]
2. Evaluate each part:
   \[
   \int_0^b (e^{-x} - a) \, dx = \left[ -e^{-x} \right]_0^b - ab = (1 - e^{-b}) - ab
   \]
   \[
   \int_b^1 (a - e^{-x}) \, dx = a(1 - b) + \left[ -e^{-x} \right]_b^1 = a(1 - b) + (e^{-b} - e^{-1})
   \]
3. Combine the results:
   \[
   \int_0^1 |e^{-x} - a| \, dx = (1 - e^{-b} - ab) + (a(1 - b) + e^{-b} - e^{-1})
   \]
   \[
   = 1 - e^{-b} - ab + a(1 - b) + e^{-b} - e^{-1}
   \]
   \[
   = 1 - ab + a(1 - b) - e^{-1}
   \]
   \[
   = 1 - ab + a - ab - e^{-1}
   \]
   \[
   = 1 + e^{-1} - a + 2a \ln a
   \]
4. Find the critical points by taking the derivative and setting it to zero:
   \[
   \frac{d}{da} \left( 1 + e^{-1} - a + 2a \ln a \right) = -1 + 2 \ln a + 2 = 2 \ln a + 1
   \]
   \[
   2 \ln a + 1 = 0 \implies \ln a = -\frac{1}{2} \implies a = e^{-\frac{1}{2}}
   \]
5. Evaluate the function at \( a = e^{-\frac{1}{2}} \):
   \[
   1 + e^{-1} - 2e^{-\frac{1}{2}}
   \]

### Conclusion
Comparing the values from all cases:
- Case 1: \( e^{-1} \)
- Case 2: \( 1 - 2e^{-1} \)
- Case 3: \( 1 + e^{-1} - 2e^{-\frac{1}{2}} \)

The minimum value is \( 1 - 2e^{-1} \).

The final answer is \( \boxed{ 1 - 2e^{-1} } \)