Search Results for ""

The Fourier transform of the delta function is given by F_x[delta(x-x_0)](k) = int_(-infty)^inftydelta(x-x_0)e^(-2piikx)dx (1) = e^(-2piikx_0). (2)

The Fourier transform of the Heaviside step function H(x) is given by F_x[H(x)](k) = int_(-infty)^inftye^(-2piikx)H(x)dx (1) = 1/2[delta(k)-i/(pik)], (2) where delta(k) is ...

The Fourier transform of the generalized function 1/x is given by F_x(-PV1/(pix))(k) = -1/piPVint_(-infty)^infty(e^(-2piikx))/xdx (1) = ...

F_x[1/pi(1/2Gamma)/((x-x_0)^2+(1/2Gamma)^2)](k)=e^(-2piikx_0-Gammapi|k|). This transform arises in the computation of the characteristic function of the Cauchy distribution.

Let R(x) be the ramp function, then the Fourier transform of R(x) is given by F_x[R(x)](k) = int_(-infty)^inftye^(-2piikx)R(x)dx (1) = i/(4pi)delta^'(k)-1/(4pi^2k^2), (2) ...

Let Pi(x) be the rectangle function, then the Fourier transform is F_x[Pi(x)](k)=sinc(pik), where sinc(x) is the sinc function.

The interesting function defined by the definite integral G(x)=int_0^xsin(tsint)dt, illustrated above (Glasser 1990). The integral cannot be done in closed form, but has a ...

G = int_0^infty(e^(-u))/(1+u)du (1) = -eEi(-1) (2) = 0.596347362... (3) (OEIS A073003), where Ei(x) is the exponential integral. Stieltjes showed it has the continued ...

A substitution which can be used to transform integrals involving square roots into a more tractable form. form substitution sqrt(x^2+a^2) x=asinhu sqrt(x^2-a^2) x=acoshu

A polynomial admitting a multiplicative inverse. In the polynomial ring R[x], where R is an integral domain, the invertible polynomials are precisely the constant polynomials ...