Prove that the integral of $fracsin(x)x$ from $0$ to $infty$ is $fracpi2$

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

${\int }_0^{\mathrm{\infty }}\frac{\sin (x)}{x}dx=\frac{\pi }{2}$

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

${\int }_0^t\frac{\sin (x)}{x}dx={\int }_0^1\frac{\sin (tu)}{tu}tdu={\int }_0^1\frac{\sin (tu)}{u}du$

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

${\int }_0^{\mathrm{\infty }}\frac{\sin (x)}{x}dx=\frac{\pi }{2}$

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

$in{t}_0^{i}nftysin(x^2),dx=sqrtfracpi2$

The Fresnel sine integral is defined as:

$S(t)=in{t}_0^{t}sin(x^2),dx$

As $toinfty$, it has been shown that:

$S(t)osqrtfracpi2$

Thus:

$in{t}_0^{i}nftysin(x^2),dx=sqrtfracpi2$

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

$fracddxsin(u)=cos(u)cdotfracdudx$

Let $u=x^2$, so:

$fracddxsin(x^2)=cos(x^2)cdot2x$

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