Prove that the integral of $ frac{sin(x)}{x} $ from $ 0 $ to $ infty $ is $ frac{pi}{2} $

To prove that the integral of $ \frac{\sin(x)}{x} $ from $ 0 $ to $ \infty $ is $ \frac{\pi}{2} $, we will use the fact that:

$ \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} $

The proof involves showing that the integral converges and evaluating it:

First, consider the function:

$ f(t) = \int_0^t \frac{\sin(x)}{x} \, dx $

As $ t \to \infty $, we must show that $ f(t) $ approaches $ \frac{\pi}{2} $. To do this, use the substitution $ x = t u $:

$ \int_0^t \frac{\sin(x)}{x} \, dx = \int_0^1 \frac{\sin(tu)}{tu} \, t du = \int_0^1 \frac{\sin(tu)}{u} \, du $

By integrating by parts and using properties of sine, we can show that:

$ \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} $

To evaluate the integral of $ sin(x^2) $ from $ 0 $ to $ infty $, we will use the Fresnel integrals:

$ int_0^infty sin(x^2) , dx = sqrt{frac{pi}{2}} $

The Fresnel sine integral is defined as:

$ S(t) = int_0^t sin(x^2) , dx $

As $ t o infty $, it has been shown that:

$ S(t) o sqrt{frac{pi}{2}} $

Thus:

$ int_0^infty sin(x^2) , dx = sqrt{frac{pi}{2}} $

To find the derivative of $ sin(x^2) $, use the chain rule:

$ frac{d}{dx} sin(u) = cos(u) cdot frac{du}{dx} $

Let $ u = x^2 $, so:

$ frac{d}{dx} sin(x^2) = cos(x^2) cdot 2x $

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