Solution

The convolution is

(f * g) (x) = \int_{- \infty}^{\infty} f (x - y) g (y) d y .

We claim that the

n

-fold convolution is

F_{n} (x) = ⎩ ⎨ ⎧ \frac{x ^{n - 1}}{( n - 1 )!} e^{- x}, 0, x \geq 0, x < 0.

This is true for

n = 1

. If it holds for

n - 1

and

x \geq 0

, then

F_{n} (x) = \int_{0}^{x} e^{- (x - y)} \frac{y ^{n - 2} e ^{- y}}{( n - 2 )!} d y = \frac{e ^{- x}}{( n - 2 )!} \int_{0}^{x} y^{n - 2} d y = \frac{x ^{n - 1} e ^{- x}}{( n - 1 )!},

and the convolution vanishes for

x < 0

The Fourier transform is

F_{n} (k) = \frac{1}{( n - 1 )!} \int_{0}^{\infty} x^{n - 1} e^{- (1 + ik) x} d x = \frac{1}{( 1 + ik ) ^{n}} .

f * g (k) = f (k) g (k) .

Indeed, Fubini and

u = x - y

give

f * g (k) = \int \int e^{- ik x} f (x - y) g (y) d y d x = \int \int e^{- ik (u + y)} f (u) g (y) d u d y = f (k) g (k) .

Since

F (k) = (1 + ik)^{- 1}

, induction immediately verifies

F_{n} = (F)^{n} = (1 + ik)^{- n} .

g (x) = \overline{f (- x)}

. Then

(f * g) (0) = \int_{- \infty}^{\infty} ∣ f (x) ∣^{2} d x, g (k) = \overline{f (k)} .

Using Fourier inversion at zero, which follows from the supplied delta identity,

(f * g) (0) = \frac{1}{2 π} \int_{- \infty}^{\infty} f * g (k) d k,

and the convolution theorem gives

\int_{- \infty}^{\infty} ∣ f (x) ∣^{2} d x = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ f (k) ∣^{2} d k .

Apply this to

F_{n + 1}

. Since

\int_{0}^{\infty} ∣ F_{n + 1} (x) ∣^{2} d x = \frac{1}{( n ! ) ^{2}} \int_{0}^{\infty} x^{2 n} e^{- 2 x} d x = \frac{( 2 n )!}{2 ^{2 n + 1} ( n ! ) ^{2}},

\int_{- \infty}^{\infty} \frac{d k}{( 1 + k ^{2} ) ^{n + 1}} = \frac{π ( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} = \frac{π}{4 ^{n}} (n 2 n) .

Solved by gpt-5.6-sol high.

 Ancestors (10)