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(1) where H_n(x) is a Hermite polynomial (Watson 1933; Erdélyi 1938; Szegö 1975, p. 380). The generating function ...

The inverse of the Laplace transform F(t) = L^(-1)[f(s)] (1) = 1/(2pii)int_(gamma-iinfty)^(gamma+iinfty)e^(st)f(s)ds (2) f(s) = L[F(t)] (3) = int_0^inftyF(t)e^(-st)dt. (4)

The so-called generalized Fourier integral is a pair of integrals--a "lower Fourier integral" and an "upper Fourier integral"--which allow certain complex-valued functions f ...

Let alpha(x) be a monotone increasing function and define an interval I=(x_1,x_2). Then define the nonnegative function U(I)=alpha(x_2)-alpha(x_1). The Lebesgue integral with ...

Integration under the integral sign is the use of the identity int_a^bdxint_(alpha_0)^alphaf(x,alpha)dalpha=int_(alpha_0)^alphadalphaint_a^bf(x,alpha)dx (1) to compute an ...

If Omega subset= C is a domain and phi:Omega->C is a one-to-one analytic function, then phi(Omega) is a domain, and area(phi(Omega))=int_Omega|phi^'(z)|^2dxdy (Krantz 1999, ...

P_n(cosalpha)=(sqrt(2))/piint_0^alpha(cos[(n+1/2)phi])/(sqrt(cosphi-cosalpha))dphi, where P_n(x) is a Legendre polynomial.

The integral of 1/r over the unit disk U is given by intint_(U)(dA)/r = intint_(U)(dxdy)/(sqrt(x^2+y^2)) (1) = int_0^(2pi)int_0^1(rdrdtheta)/r (2) = 2piint_0^1dr (3) = 2pi. ...

Let the elliptic modulus k satisfy 0<k^2<1, and the Jacobi amplitude be given by phi=amu with -pi/2<phi<pi/2. The incomplete elliptic integral of the first kind is then ...

A_m(lambda)=int_(-infty)^inftycos[1/2mphi(t)-lambdat]dt, (1) where the function phi(t)=4tan^(-1)(e^t)-pi (2) describes the motion along the pendulum separatrix. Chirikov ...