Consider an integrable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $af(a)+bf(b)=0$ when $ab=1$. Find the value of the following integration:
$$ \int_{0}^{\infty} f(x) \,dx $$
1. Given the condition \( af(a) + bf(b) = 0 \) when \( ab = 1 \), we can rewrite this as:
   \[
   af(a) + \frac{1}{a} f\left(\frac{1}{a}\right) = 0
   \]
   This implies:
   \[
   a f(a) = -\frac{1}{a} f\left(\frac{1}{a}\right)
   \]
   Multiplying both sides by \( a \), we get:
   \[
   a^2 f(a) = -f\left(\frac{1}{a}\right)
   \]
   Therefore:
   \[
   f\left(\frac{1}{a}\right) = -a^2 f(a)
   \]

2. We need to evaluate the integral:
   \[
   \int_{0}^{\infty} f(x) \, dx
   \]

3. To do this, we perform a substitution. Let \( x = \frac{1}{t} \). Then \( dx = -\frac{1}{t^2} dt \). The limits of integration change as follows:
   - When \( x \to 0 \), \( t \to \infty \)
   - When \( x \to \infty \), \( t \to 0 \)

   Therefore, the integral becomes:
   \[
   \int_{0}^{\infty} f(x) \, dx = \int_{\infty}^{0} f\left(\frac{1}{t}\right) \left(-\frac{1}{t^2}\right) dt
   \]

4. Simplifying the integral, we get:
   \[
   \int_{0}^{\infty} f(x) \, dx = \int_{\infty}^{0} -\frac{1}{t^2} f\left(\frac{1}{t}\right) dt
   \]
   Changing the limits of integration back to \( 0 \) to \( \infty \), we have:
   \[
   \int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} \frac{1}{t^2} f\left(\frac{1}{t}\right) dt
   \]

5. Using the relationship \( f\left(\frac{1}{t}\right) = -t^2 f(t) \), we substitute:
   \[
   \int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} \frac{1}{t^2} \left(-t^2 f(t)\right) dt
   \]
   Simplifying, we get:
   \[
   \int_{0}^{\infty} f(x) \, dx = \int_{0}^{\infty} -f(t) dt
   \]

6. Therefore:
   \[
   \int_{0}^{\infty} f(x) \, dx = -\int_{0}^{\infty} f(t) dt
   \]

7. Adding \(\int_{0}^{\infty} f(x) \, dx\) to both sides, we get:
   \[
   2 \int_{0}^{\infty} f(x) \, dx = 0
   \]

8. Dividing both sides by 2, we find:
   \[
   \int_{0}^{\infty} f(x) \, dx = 0
   \]

The final answer is \(\boxed{0}\).