Q8Advanced Engineering Mathematics

The problem demands a rigorous evaluation of the Laplace transform of a complex trigonometric fraction, followed by the deduction of a specific improper integral. The Laplace transform is exceptionally powerful for evaluating integrals from 0 to infinity that are otherwise intractable using standard real analysis techniques.

1. Finding the Laplace Transform

We must evaluate $L\left\{\frac{{\sin }^{2}t}{t}\right\}$. The function has $t$ in the denominator, which immediately indicates that we must use the "Division by $t$" property: $L\left\{\frac{f(t)}{t}\right\}={\int }_s^{\infty }F(u)du$, where $F(s)=L\{f(t)\}$.

First, we find the transform of the numerator $f(t)={\sin }^{2}t$. We cannot transform a squared trigonometric function directly. We must use the double-angle power-reduction identity to linearize it: ${\sin }^{2}t=\frac{1-\cos (2t)}{2}$.

Now, we apply the linearity property of the Laplace transform to this expression:

Next, we apply the Division by $t$ integration theorem:

We integrate the two terms. The first is a simple natural logarithm. The second term requires a u-substitution (let $w=u^2+4$, $dw=2udu$, so $\int \frac{u}{u^2+4}du=\frac{1}{2}\ln (u^2+4)$).

Using logarithmic properties ($\frac{1}{2}\ln (A)=\ln (A^{1/2})$ and $\ln (A)-\ln (B)=\ln (A/B)$), we combine the terms into a single compact logarithmic expression:

To evaluate the upper limit as $u\to \infty$, we analyze the behavior of the argument. We divide numerator and denominator by $u$: $\frac{1}{\sqrt{1+4/u^2}}$. As $u$ approaches infinity, $4/u^2$ approaches zero, leaving $\frac{1}{\sqrt{1}}=1$. The natural logarithm of 1 is exactly 0. Thus, the upper bound evaluation is 0.

We eliminate the negative sign by inverting the argument inside the logarithm:

This completes the first part of the proof.

2. Deduction of the Improper Integral

The second part of the question asks us to deduce the value of ${\int }_0^{\infty }\frac{{\sin }^{2}t}{t^2}dt$. This requires another application of Laplace theory.

We know by definition that $L\{g(t)\}={\int }_0^{\infty }{e}^{-st}g(t)dt$. If we take the mathematical limit as $s\to 0$ on both sides (assuming the integral is convergent), we obtain ${\lim }_{s\to 0}L\{g(t)\}={\int }_0^{\infty }g(t)dt$.

Let $g(t)=\frac{{\sin }^{2}t}{t^2}$. We need its Laplace transform. We apply the Division by $t$ property a second time to our previously derived result:

Evaluating this new integral from 0 to infinity (taking limit $s\to 0$) requires advanced integration by parts. Let $U=\ln (1+4/u^2)$ and $dV=du$. Then $dU=\frac{-8/u^3}{1+4/u^2}du=\frac{-8}{u(u^2+4)}du$ and $V=u$.

We evaluate this from $s\to 0$ to $\infty$. As $u\to \infty$, $u\ln (1+4/u^2)\to 0$ (via L'Hopital's rule) and $\arctan (\infty )=\pi /2$. As $u\to 0$, $u\ln (1+4/u^2)\to 0$ and $\arctan (0)=0$.

The definite integral evaluates to $0+4(\pi /2)-(0+0)=2\pi$.

Remembering the $1/4$ factor in front of the integral: $I=\frac{1}{4}\times 2\pi =\frac{\pi }{2}$.

Therefore, it is rigorously proved that ${\int }_0^{\infty }\frac{{\sin }^{2}t}{t^2}dt=\frac{\pi }{2}$.