Find more Fourier integrals
$
\def \half {\frac{1}{2}}
\def \hieruit {\quad \Longrightarrow \quad}
\def \sinc {\operatorname{sinc}}
$
High on the wish list of people who find Laplace transforms interesting is the
Laplace transform of a sine function:
$$
\int_0^{\infty} e^{-s x} \sin(x) \, dx = \mbox{??}
$$
We proceed by integrating by parts:
$$
\int_0^\infty e^{-s x} \sin(x) \, dx =
\int_0^\infty e^{-s x} d\left[ - \cos(x) \right] =
$$ $$
- \left[ e^{-s x} \cos(x) \right]_0^\infty
+ \int_0^\infty \cos(x) \, d \left[ e^{-s x} \right] =
1 - s \int_0^\infty e^{-s x} \cos(x) \, dx
$$
Feels good. Let's do it again:
$$
1 - s \int_0^\infty e^{-s x} \cos(x) \, dx =
1 - s \int_0^\infty e^{-s x} d \left[ \sin(x) \right] =
$$ $$
1 - s \left[ e^{-s x} \sin(x) \right]_0^\infty
+ s \, \int_0^\infty \sin(x) \, d \left[ e^{-s x} \right] =
$$ $$
1 - s^2 \, \int_0^\infty e^{-s x} \sin(x) \, d x =
\int_0^\infty e^{-s x} \sin(x) \, d x \hieruit
$$
Theorem. Laplace transform of sine:
$$
\int_0^\infty e^{-s x} \sin(x) \, dx = \frac{1}{1 + s^2}
$$
Theorem. Laplace transform of cosine:
$$
\int_0^\infty e^{-s x} \cos(x) \, dx = \frac{s}{1 + s^2}
$$
Proof.
$$
\int_0^\infty e^{-s x} \cos(x) \, dx = \int_0^\infty e^{-s x} \, d \sin(x) =
\left[e^{-s x}\sin(x)\right]_0^\infty - \int_0^\infty \sin(x)\,d e^{-s x}
$$ $$
= 0 + s \int_0^\infty e^{-s x} \sin(x) \, dx = \frac{s}{1 + s^2}
$$
Theorem. Fourier integral of (normed) decay function:
$$
\int_{-\infty}^{+\infty} \half e^{-|x|} \cos(y x) \,dx = \frac{1}{1 + y^2}
$$
Proof.
$$
\int_0^\infty e^{-s x} \cos(x) \,dx = \frac{s}{1 + s^2} \hieruit
\int_{-\infty}^{+\infty} \half e^{-|x|} \cos(y x) \,dx =
$$ $$
1/y \, \int_0^\infty e^{-(y x)/y} \cos(y x) \,d(y x) =
1/y \, \frac{1/y}{1 + (1/y)^2} = \frac{1}{1 + y^2}
$$
Theorem. Half area of sinc function:
$$
\int_0^\infty \sinc(x)\,dx = \frac{\pi}{2}
\hieruit \mbox{normed} \: = \frac{1}{\pi} \sinc(x)
$$
Proof.
$$
\int_0^\infty \left[ \int_0^\infty e^{-s x} \sin(x) \, dx \right] \, ds =
\int_0^\infty \left[ \int_0^\infty e^{-s x} \sin(x) \, ds \right] \, dx
$$
Because iterated limits in the real world do always commute. Now we have a left
hand side and a right hand side. The left hand side, according to the above, is
equal to:
$$
\int_0^\infty \left[ \int_0^\infty e^{-s x} \sin(x) \, dx \right] \, ds =
\int_0^\infty \frac{ds}{1+s^2} = \left[ \arctan(s) \right]_0^\infty =
\frac{\pi}{2}
$$
The right hand side, on the other hand:
$$
\int_0^\infty \left[ \int_0^\infty e^{-s x} \sin(x) \, ds \right] \, dx =
\int_0^\infty \sin(x) \left[ \int_0^\infty e^{-s x} \, ds \right] \, dx =
$$ $$
\int_0^\infty \sin(x) \frac{1}{x} \left[ - e^{-y} \right]_0^\infty \, dx =
\int_0^\infty \frac{\sin(x)}{x} \, dx \qquad \mbox{Q.E.D.}
$$
Theorem. Fourier integral of sinc function:
$$
\int_{-\infty}^{+\infty} \sinc(x) \cos(y x) \,dx = A(y)
$$
Where:
$$
A(y) = \left\{ \begin{array}{lll}
0 & \mbox{for} & y < -1 \\
\pi & \mbox{for} & -1 < y < +1 \\
0 & \mbox{for} & +1 < y
\end{array} \right.
$$
Proof.
$$
2\,\sin(\alpha)\cos(\beta) = \sin(\alpha+\beta) + \sin(\alpha-\beta) \hieruit
$$ $$
\int_{-\infty}^{+\infty} \frac{\sin(x)\cos(y x)}{x} \,dx =
\half \int_{-\infty}^{+\infty} \frac{\sin(x+y x)}{x} \,dx +
\half \int_{-\infty}^{+\infty} \frac{\sin(x-y x)}{x} \,dx
$$ $$
= \half \int_{-\infty}^{+\infty} \frac{\sin((1+y)x)}{(1+y)x} (1+y) \,dx +
\half \int_{-\infty}^{+\infty} \frac{\sin((1-y)x)}{(1-y)x} (1-y) \,dx
$$
Now let $u = (1+y)x$ and $v = (1-y)x$ . Then:
$$
\int_{-\infty}^{+\infty} \frac{\sin((1+y)x)}{(1+y)x} (1+y) \,dx =
\left\{ \begin{array}{ll}
+ \int_{-\infty}^{+\infty} \sinc(u)\,du & \mbox{for} \quad 1+y > 0 \\
- \int_{-\infty}^{+\infty} \sinc(u)\,du & \mbox{for} \quad 1+y < 0
\end{array} \right.
$$
$$
\int_{-\infty}^{+\infty} \frac{\sin((1-y)x)}{(1-y)x} (1-y) \,dx =
\left\{ \begin{array}{ll}
+ \int_{-\infty}^{+\infty} \sinc(v)\,dv & \mbox{for} \quad 1-y > 0 \\
- \int_{-\infty}^{+\infty} \sinc(v)\,dv & \mbox{for} \quad 1-y < 0
\end{array} \right.
$$
Where
$\int_{-\infty}^{+\infty} \sinc(u)\,du = 2\,\int_0^\infty \sinc(v)\,dv
= 2.\half\pi = \pi$. Summarizing:
|
$\int_{-\infty}^{+\infty} \sinc(u)\,du$ |
$\int_{-\infty}^{+\infty} \sinc(v)\,dv$ |
$\int_{-\infty}^{+\infty} \sinc(x) \cos(y x) \,dx$ |
| $y < -1$ | $-\pi$ |
$+\pi$ | $0$ |
$-1 < y < +1$ | $+\pi$ |
$+\pi$ | $\pi$ |
$+1 < y$ | $+\pi$ |
$-\pi$ | $0$ |
Theorem. Parceval's theorem for sinc function:
$$
\int_{-\infty}^{+\infty} \sinc^2(x)\,dx = \pi
\hieruit \mbox{normed} \: = \frac{1}{\pi} \sinc^2(x)
$$
Proof. The integral of the square of the Fourier transform of a function
is equal to the integral of the square of the function itself:
$$
\int_{-\infty}^{+\infty} f^2(x)\, dx =
\frac{1}{2\pi} \int_{-\infty}^{+\infty} A^2(y)\, dy
$$ $$
\hieruit \int_{-\infty}^{+\infty} \sinc^2(x)\,dx =
\frac{1}{2\pi} \int_{-\infty}^{+\infty} A^2(y)\,dy =
\frac{1}{2\pi} \int_{-1}^{+1} \pi^2\,dx = \frac{2\pi^2}{2\pi} = \pi
$$
Because:
$$
A(y) = \left\{ \begin{array}{lll}
0 & \mbox{for} & y < -1 \\
\pi & \mbox{for} & -1 < y < +1 \\
0 & \mbox{for} & +1 < y
\end{array} \right. \qquad \mbox{Q.E.D.}
$$