Download The Fourier transform of e

The Fourier transform of e−ax
2
R∞
2
We will also need the value of the integral I = −∞ e−x dx to which
end we consider
Z ∞
Z ∞
Z ∞Z ∞
2
2
2
−x2
−y 2
I =
e
dx ·
e
dy =
e−(x +y ) dy dx
−∞
−∞ −∞
Z−∞
Z
2
2
=
e−(x +y ) dA.
R2
Introduction
2
Let a > 0 be constant. We define a function fa (x) by fa (x) = e−ax and
denote by fˆa (w) the Fourier transform of fa (x). We wish to show that
fˆa (w) = F (fa )(w) =
2
√1 e−w /(4a)
2a
for all w ∈ R.
The proof consists of three parts,
1. ancillary results
2. the derivation of an ordinary differential equation for fˆa (w), as
well as a suitable initial condition to determine a uniqe solution of
the ODE
3. the resolution of the initial value problem (i.e. the ODE and the
initial condition)
We will only require tools already presented in prior mathematics courses
at NTNU.
Ancillary results
Using the substitution u = −ax2 we get du = −2ax dx and x dx =
du/(−2a). Consequently,
Z
Z
−ax2
1
1 u
1 −ax2
xe
dx = eu −2a
du = − 2a
e + C = − 2a
e
+ C.
1
Using polar coordinates x = r cos θ and y = sin θ, the above integral
may be rewritten as
Z 2π Z ∞
Z ∞
2
2
−r2
I =
e
r dr dθ = 2π
re−r dr.
0
0
0
Thus
2 ∞
I 2 = 2π · − 21 e−r 0 = π
√
and we finally get I = π.
A differential equation for fˆa (w)
Direct insertion into the definition of the Fourier transform yields
Z ∞
1
2
ˆ
√
fa (w) =
e−ax e−iwx dx.
2π −∞
We note in particular that
Z ∞
1
2
fˆa (0) = √
e−ax dx.
2π −∞
√
√
Using the substitution u = a x we find du = a dx from which dx =
√
du/ a. Consequently,
Z ∞
Z ∞
1
1
2
−u2 √1
ˆ
du = √
fa (0) = √
e
e−u du
a
2π −∞
2πa −∞
(1)
√
1
1
· π=√ .
=√
2πa
2a
2
We incidentally remark that this result is consistent with the desired
end result we wish to prove. Differentiating fˆa (w) with respect to the
independent variable w further gives
Z ∞
dfˆa
d 1
2
√
(w) =
e−ax e−iwx dx
dw
dw 2π −∞
Z ∞
1
∂ −ax2 −iwx
(2)
=√
(e
e
) dx
∂w
2π −∞
Z ∞
i
2
= −√
xe−ax e−iwx dx.
2π −∞
Integrating equation (2) by parts now gives
Z
i
dfˆa
iw ∞ −ax2 −iwx −ax2 −iwx ∞
1
e
e
dx
(w) = − √
e
]−∞ −
[− e
dw
{z
} 2a −∞
2π | 2a
Z ∞ =0
w 1
w
2
√
=−
e−ax e−iwx dx = − fˆa (w),
2a 2π −∞
2a
We determine the value of C by means of the initial value as fˆa (0) =
C = √12a . In other words,
fˆa (w) =
2
√1 e−w /(4a)
2a
as we wanted to prove.
Concluding remarks
It is particularly noteworthy that the Fourier transform of fa (x) is another function of the same general form as the function itself. That is,
2
fˆa (w) = e−kw in which k is a positive constant. This is highly remarkable and does not hold in the general case. On the other hand, other
functions do exist for which the Fourier transform is of the same form
as the original function, meaning that fa (x) is not entirely exceptional
in this respect.
the latter equality following from the definition of the Fourier transform
of fa (x).
Resolution of the initial value problem for fˆa (w)
Combining this with the result (1) we find
dfˆa
w
= − fˆa ,
dw
2a
√
fˆa (0) = 1/ 2a.
This is a linear, first order separable ordinary differential equation and
initial value to determine fˆa . The process of separation then gives
dfˆa
w
= − dw
ˆ
2a
fa
2
from which anti-differentiation yields ln fˆa (w) = − w4a + K. Here, K is
an arbitrary constant of integration. Consequently
2
fˆa (w) = C e−w /(4a) ,
3
C = eK .
4