f (s) = \int_{- \infty}^{\infty} f (x) e^{- i s x} d x .

(1)

For

f, f^{'} \in L^{1} (R)

with the usual local absolute-continuity hypothesis, the Fourier transform of a derivative is

f^{'} (s) = i s f (s) .

(2)

For example, this applies when

f

C^{1}

and both displayed functions are integrable. These assumptions make integration by parts valid and the endpoint terms vanish.

ii

 0  0

Solution

 0  0

Split the integral at the origin and reflect the negative half:

e^{- a ∣ x ∣} (s) = \int_{0}^{\infty} e^{- (a + i s) x} d x + \int_{0}^{\infty} e^{- (a - i s) x} d x = \frac{1}{a + i s} + \frac{1}{a - i s} = \frac{2 a}{a ^{2} + s ^{2}} .

(1)

Thus the Fourier transform of a two-sided exponential is

e^{- a ∣ x ∣} (s) = \frac{2 a}{a ^{2} + s ^{2}} (a > 0) .

(2)

iii

 0  0

Solution

 0  0

With this convention, the Fourier inversion theorem states that if

f, f \in L^{1} (R)

, then

f (x) = \frac{1}{2 π} \int_{R} f (s) e^{i s x} d s

(1)

almost everywhere and at every continuity point of

f

. Apply this to the continuous function in part (ii):

e^{- a ∣ x ∣} = \frac{a}{π} \int_{R} \frac{e ^{i s x}}{a ^{2} + s ^{2}} d s .

(2)

The denominator is even. Replacing

x

- t

and renaming the integration variable gives

F (\frac{1}{a ^{2} + x ^{2}}) (t) = \frac{π}{a} e^{- a ∣ t ∣} .

(3)

Both integrable members of this transform pair satisfy the inversion hypotheses.

iv

 0  0

Solution

 0  0

Complete the square:

x^{2} + 2 x + 2 = (x + 1)^{2} + 1

. In the Fourier transform, the substitution

u = x + 1

gives

F (\frac{1}{x ^{2} + 2 x + 2}) (s) = π e^{- ∣ s ∣} e^{i s} .

(1)

The phase

e^{i s}

is important. Taking the real part at

s = a

gives the cosine integral, since changing the sign in the exponential does not change its real part:

\int_{- \infty}^{\infty} \frac{cos ( a x )}{x ^{2} + 2 x + 2} d x = π e^{- ∣ a ∣} cos a (a \in R) .

(2)

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