Find the form of Parseval’s theorem(12.24)for sine transforms (12.14)and for cosine transforms(12.15).

We have given a cosine and sine transform

Fourier Cosine transform: we define$f_c\left(x\right)$and$g_c\left(\alpha \right)$a pair ofFourier

Cosine transformsrepresentingeven functions, by the equations

$$\begin{gathered} f_c\left(x\right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{g}_c\left(x\right)\cos \left(\alpha x\right)d\alpha \\ {g}_c\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{f}_c\left(x\right)\cos \left(\alpha x\right)dx \end{gathered}$$

Fourier Sine transform: we define$f_s\left(x\right)$and$g_s\left(\alpha \right)$a pair ofFourier

Sine transformsrepresentingodd functions, by the equations

$$\begin{gathered} f_s\left(x\right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{g}_s\left(\alpha \right)\sin \left(\alpha x\right)d\alpha \\ {g}_s\left(x\right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{f}_s\left(x\right)\sin \left(\alpha x\right)dx \end{gathered}$$

Let us start with the cosine transform

$g_1\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)\cos \left(\alpha x\right)dx$

Multiply both sides with $g_2\left(\alpha \right)$and integrate from 0 to infinity

$$\begin{gathered} \underset{0}{\overset{\infty }{\int }}{g}_1\left(\alpha \right)g_2\left(\alpha \right)d\alpha =\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)\cos \left(\alpha x\right)dx\underset{0}{\overset{\infty }{\int }}{g}_2\left(\alpha \right)d\alpha \\ =\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)\left[\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{g}_2\left(x\right)\cos \left(\alpha x\right)d\alpha \right]dx \\ =\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)f_2\left(x\right)dx \end{gathered}$$

By setting$f_1=f_2=f$ and$g_1=g_2=g$ we get

$\underset{0}{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{0}{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$

Let us start with the sine transform

$g_1\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)\sin \left(\alpha x\right)dx$

Multiply both sides with$g_2\left(\alpha \right)$and integrate from 0 to infinity

$$\begin{gathered} \underset{0}{\overset{\infty }{\int }}{g}_1\left(\alpha \right)g_2\left(\alpha \right)d\alpha =\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)\sin \left(\alpha x\right)dx\underset{0}{\overset{\infty }{\int }}{g}_2\left(\alpha \right)d\alpha \\ =\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)\left[\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}{g}_2\left(x\right)\sin \left(\alpha x\right)d\alpha \right]dx \\ =\underset{0}{\overset{\infty }{\int }}{f}_1\left(x\right)f_2\left(x\right)dx \end{gathered}$$

By setting$f_1=f_2=f$and$g_1=g_2=g$we get

$\underset{0}{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{0}{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$.

Therefore the form of Parseval’s theorem is$\underset{0}{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{0}{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$