Q12Advanced Engineering Mathematics

The problem requires the calculation of the Fourier Transform for a function $f(x)$ defined as $1-x^2$ over the finite interval $[-1,1]$, and zero elsewhere. The standard mathematical definition of the Fourier transform $F(s)$ is given by the infinite integral $F(s)={\int }_{-\infty }^{\infty }f(x)e^{isx}dx$. Because our specific function is identically zero outside the interval $[-1,1]$, the infinite integral securely collapses to a definite integral with finite bounds.

1. Using Symmetry Properties

We can vastly simplify the integration by leveraging the parity (symmetry) of the function. Using Euler's formula, $e^{isx}=\cos (sx)+i\sin (sx)$. Since $1-x^2$ is an even algebraic function, multiplying it by the odd trigonometric function $\sin (sx)$ results in an odd function. The definite integral of an odd function over symmetric limits $[-1,1]$ is exactly zero. Multiplying by the even function $\cos (sx)$ yields an even function, whose integral over $[-1,1]$ is twice the integral from 0 to 1.

2. Applying Integration by Parts

We must now evaluate the integral $I={\int }_0^1(1-x^2)\cos (sx)dx$. This requires the technique of integration by parts, formula: $\int udv=uv-\int vdu$. We strategically choose $u=(1-x^2)$ so that it differentiates to zero eventually, and $dv=\cos (sx)dx$. Thus, $du=-2xdx$ and $v=\frac{\sin (sx)}{s}$.

The first term evaluates to zero at both upper bound ($1-1=0$) and lower bound ($\sin (0)=0$).

We must apply integration by parts a second time for $\int x\sin (sx)dx$. Let $u=x$, $dv=\sin (sx)dx$, so $du=dx$, $v=-\frac{\cos (sx)}{s}$.

Multiplying the expression by the coefficient 2 we derived earlier from symmetry, the final Fourier transform is: