To prove the given equality:

$\sum_{n=0}^\infty \int_0^1 x^n \sin(\pi x) \, dx = \int_0^\pi \frac{\sin x}{x} \, dx,$

we will break the proof into several steps:

Step 1: Swap the summation and integral

The summation and integral can be swapped under conditions of uniform convergence, which is satisfied in this case. Thus, we rewrite the left-hand side as: $\int_0^1 \left( \sum_{n=0}^\infty x^n \right) \sin(\pi x) \, dx.$

Step 2: Evaluate the infinite geometric series

The series $\sum_{n=0}^\infty x^n$ converges to $\frac{1}{1 - x}$ for $x \in [0, 1)$. Substituting this result, we get: $\int_0^1 \frac{\sin(\pi x)}{1 - x} \, dx.$

Step 3: Substitute $x = \frac{t}{\pi}$

To simplify the integration, let $x = \frac{t}{\pi}$. Then $dx = \frac{dt}{\pi}$, and the limits of integration change as follows:

• When $x = 0$, $t = 0$,
• When $x = 1$, $t = \pi$.

Substituting, the integral becomes: $\int_0^1 \frac{\sin(\pi x)}{1 - x} \, dx = \int_0^\pi \frac{\sin t}{t} \, dt.$

Conclusion

Thus, the left-hand side simplifies exactly to the right-hand side: $\sum_{n=0}^\infty \int_0^1 x^n \sin(\pi x) \, dx = \int_0^\pi \frac{\sin x}{x} \, dx.$

This completes the proof.

Let me know if you need further details or explanations!

Related Questions:

1. What conditions must hold for the summation and integral to be swapped?
2. How does the convergence of the geometric series $\sum_{n=0}^\infty x^n$ ensure the proof works?
3. Could this technique be applied to similar integrals with different trigonometric functions?
4. What happens if we change the limits of integration on the right-hand side?
5. How does the substitution $x = \frac{t}{\pi}$ simplify the problem?

Tip: When working with infinite series inside integrals, always verify uniform convergence to ensure valid swapping of operations!