We are tasked with showing that:

$e^{-x} \sin(x) > e^{-x^2} \sin(x) \quad \text{for} \quad x \in (0, 1)$

Step 1: Simplify the inequality

We can divide both sides of the inequality by $\sin(x)$, since $\sin(x) > 0$ for $x \in (0, 1)$. The inequality becomes:

$e^{-x} > e^{-x^2}$

Now, we need to prove that $e^{-x} > e^{-x^2}$ for $x \in (0, 1)$.

Step 2: Take the natural logarithm of both sides

To simplify this further, take the natural logarithm of both sides:

$\ln(e^{-x}) > \ln(e^{-x^2})$

Using the properties of logarithms, this simplifies to:

$-x > -x^2$

Which is equivalent to:

$x^2 > x$

Step 3: Solve the inequality $x^2 > x$

Rearrange the inequality:

$x^2 - x > 0$

Factor the expression:

$x(x - 1) > 0$

This inequality holds when $x \in (0, 1)$, because the product of $x$ and $x - 1$ is positive in this interval.

Conclusion

Since $x^2 > x$ for $x \in (0, 1)$, we have shown that:

$e^{-x} > e^{-x^2}$

Thus, it follows that:

$e^{-x} \sin(x) > e^{-x^2} \sin(x) \quad \text{for} \quad x \in (0, 1)$

Further Exploration:

1. How does the function $f(x) = e^{-x} - e^{-x^2}$ behave for $x \in (0, 1)$?
2. What would happen if we consider $x \in (1, 2)$ instead of $x \in (0, 1)$?
3. How does the rate of decay of $e^{-x}$ compare to $e^{-x^2}$ for large $x$?
4. Can we find the exact point where $e^{-x} = e^{-x^2}$?
5. How would this inequality change if we added a constant to one side of the inequality?

Tip:

To compare functions like $e^{-x}$ and $e^{-x^2}$, consider their derivatives. The derivative of $e^{-x}$ is $-e^{-x}$, while the derivative of $e^{-x^2}$ is $-2xe^{-x^2}$. By analyzing these, you can gain more insight into their behavior in different intervals!