For real numbers $a,\ b$ with $0\leq a\leq \pi,\ a<b$, let $I(a,\ b)=\int_{a}^{b} e^{-x} \sin x\ dx.$

Determine the value of $a$ such that $\lim_{b\rightarrow \infty} I(a,\ b)=0.$
1. **Evaluate the integral using integration by parts:**
   We start with the integral \( I(a, b) = \int_{a}^{b} e^{-x} \sin x \, dx \). To solve this, we use integration by parts. Let \( u = \sin x \) and \( dv = e^{-x} dx \). Then, \( du = \cos x \, dx \) and \( v = -e^{-x} \).

   Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:
   \[
   \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int -e^{-x} \cos x \, dx
   \]
   Simplifying, we have:
   \[
   \int e^{-x} \sin x \, dx = -e^{-x} \sin x + \int e^{-x} \cos x \, dx
   \]

2. **Apply integration by parts again:**
   Now, we need to evaluate \( \int e^{-x} \cos x \, dx \). Let \( u = \cos x \) and \( dv = e^{-x} dx \). Then, \( du = -\sin x \, dx \) and \( v = -e^{-x} \).

   Using the integration by parts formula again, we get:
   \[
   \int e^{-x} \cos x \, dx = -e^{-x} \cos x - \int -e^{-x} (-\sin x) \, dx
   \]
   Simplifying, we have:
   \[
   \int e^{-x} \cos x \, dx = -e^{-x} \cos x + \int e^{-x} \sin x \, dx
   \]

3. **Combine the results:**
   Let \( I = \int e^{-x} \sin x \, dx \). From the previous steps, we have:
   \[
   I = -e^{-x} \sin x + \int e^{-x} \cos x \, dx
   \]
   and
   \[
   \int e^{-x} \cos x \, dx = -e^{-x} \cos x + I
   \]

   Substituting the second equation into the first, we get:
   \[
   I = -e^{-x} \sin x - e^{-x} \cos x + I
   \]
   Simplifying, we find:
   \[
   0 = -e^{-x} (\sin x + \cos x)
   \]
   Therefore,
   \[
   I = \frac{-1}{2} e^{-x} (\sin x + \cos x)
   \]

4. **Evaluate the definite integral:**
   Now, we evaluate the definite integral \( I(a, b) \):
   \[
   I(a, b) = \left[ \frac{-1}{2} e^{-x} (\sin x + \cos x) \right]_{a}^{b}
   \]
   This gives:
   \[
   I(a, b) = \frac{-1}{2} e^{-b} (\sin b + \cos b) - \frac{-1}{2} e^{-a} (\sin a + \cos a)
   \]
   Simplifying, we have:
   \[
   I(a, b) = \frac{1}{2} e^{-a} (\sin a + \cos a) - \frac{1}{2} e^{-b} (\sin b + \cos b)
   \]

5. **Take the limit as \( b \to \infty \):**
   We need to find \( \lim_{b \to \infty} I(a, b) \):
   \[
   \lim_{b \to \infty} I(a, b) = \lim_{b \to \infty} \left( \frac{1}{2} e^{-a} (\sin a + \cos a) - \frac{1}{2} e^{-b} (\sin b + \cos b) \right)
   \]
   Since \( e^{-b} \to 0 \) as \( b \to \infty \), we get:
   \[
   \lim_{b \to \infty} I(a, b) = \frac{1}{2} e^{-a} (\sin a + \cos a)
   \]

6. **Set the limit to zero and solve for \( a \):**
   We need \( \frac{1}{2} e^{-a} (\sin a + \cos a) = 0 \). Since \( e^{-a} \neq 0 \) for any real \( a \), we must have:
   \[
   \sin a + \cos a = 0
   \]
   Solving \( \sin a + \cos a = 0 \), we get:
   \[
   \sin a = -\cos a
   \]
   Dividing both sides by \( \cos a \), we get:
   \[
   \tan a = -1
   \]
   Therefore,
   \[
   a = \frac{3\pi}{4} + k\pi \quad \text{for integer } k
   \]
   Since \( 0 \leq a \leq \pi \), the only solution is:
   \[
   a = \frac{3\pi}{4}
   \]

The final answer is \( \boxed{\frac{3\pi}{4}} \)