Question

If F(-a^2)=0, then prove that:

i.

(1/F(D^2))sinax=x(1/d/dD[F(D^2)])sinax ii. (1/F(D^2))cosax=x(1/d/dD[F(D^2)])cosax

Answer 1

$\displaystyle\textsf{If}\,\, f(a)=0\,\, \textsf{then}\,\, D-a\,\,\textsf{is a factor}\,\,\textsf{of}\,\, f(D)\\ \textsf{So, we have}\,\, f(D) = (D-a)\phi(D)\\ \begin{aligned} \frac{1}{f(D)}e^{ax} &= \frac{1}{(D-a)(\phi(D))}e^{ax} \\&= \frac{1}{D-a}\left(\frac{1}{\phi(D)}e^{ax}\right) \\&= \frac{1}{D-a}\left(\frac{1}{\phi(a)}e^{ax}\right) \\&= \frac{1}{\phi(a)} \frac{1}{D-a}e^{ax} \\&= \frac{1}{\phi(a)} e^{ax} \int e^{-ax}\cdot e^{ax} \mathrm{d}x \\&= \frac{1}{\phi(a)} e^{ax} \int \mathrm{d}x \\&= \frac{1}{\phi(a)} xe^{ax} \end{aligned}\\ \textsf{Differentiating wrt.}\,\, D\\ f'(D) = \phi(D) + (D-a)\phi'(D)\\ \implies f'(a) = \phi(a)\\ \begin{aligned} \therefore \frac{1}{f(D)}e^{ax} &= \frac{1}{\phi(a)} xe^{ax} \\&= \frac{1}{f'(a)} xe^{ax} \\&= x \frac{1}{f'(a)} e^{ax} \\&= x \frac{1}{f'(D)} e^{ax} \end{aligned}\\ \textsf{Replace}\,\, D \,\, \textsf{with}\,\, D^2\,\, \textsf{and}\,\, a \,\, \textsf{with} \,\, ia \\ \frac{1}{f(D^2)}e^{iax} = x \frac{1}{f'(D^2)} e^{iax}\\ \frac{1}{f(D^2)}\left(\cos(ax) + i\sin(ax)\right) = x \frac{1}{f'(D^2)} \left(\cos(ax) + i\sin(ax)\right).\\ \textsf{Setting}\,\, D=ia,\,\,f(D^2) = f(-a^2) = 0\\ \textsf{Equating real and imaginary parts, we have}\\ i)\,\,\frac{1}{f(D^2)}\left(\sin(ax)\right) = x \frac{1}{f'(D^2)} \left(\sin(ax)\right)\\ \textsf{and}\\ ii)\,\,\frac{1}{f(D^2)}\left(\cos(ax)\right) = x \frac{1}{f'(D^2)} \left(\cos(ax)\right)$