The Laplace transform of a function is given by the formula. Find if

Question

The Laplace transform  of a function  is given by the formula.  Find  if

Quizplus · Accepted Answer

We can rearrange the given formula to get:

$f(t) = \frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} F(s) e^{st} ds$

where $F(s)$ is the Laplace transform.

Substituting $s=i\omega$ and $\omega=\frac{\pi}{2t}$, we get:

$f(t) = \frac{1}{\pi} \int_{-\infty}^{\infty} F(i\omega) e^{i\pi \frac{\omega}{2t}} d\omega$

Comparing with the given formula, we see that $F(i\omega) = \frac{1}{1+\omega^2}$.

Substituting $\omega=1$, we get:

$F(i) = \frac{1}{1+1^2} = \frac{1}{2}$

Therefore, using the formula for $f(t)$ and $F(s)$, we get:

$f(1) = \frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{e^s}{1+s^2} ds = \frac{1}{2\pi i} \oint_C \frac{e^z}{1+z^2} dz$

where $C$ is a contour in the complex plane including the imaginary axis and enclosing the pole at $z=i$.

By the residue theorem, this integral is equal to $2\pi i$ times the residue of the integrand at $z=i$, which is:

$\lim_{z	o i} (z-i) \frac{e^z}{1+z^2} = \frac{e^i}{2i}$

Therefore:

$f(1) = \frac{1}{2} e^i$

Taking the real part, we get:

$f(1) = \frac{1}{2} \cos 1 + \frac{i}{2} \sin 1$

The best choice is A since it is the closest to the value of $\frac{1}{2} \cos 1$.

The Laplace Transform of a Function Is Given