User talk:Timaou01

Integral formula for simple exponential function integration
Let f  be  a polynomial function differentiable on R ; α is a real or complex constant  then  the integral defined as : I=∫f(x)e(-αx )  dx  can be calculated as :

∫f(x)e(-αx ) dx=-[(f(x))/α+(f'(x))/α2 +(f(2) (x))/α3 +⋯+(f(n) (x))/α^(n+1) ]  e(-αx)                          (1)

Applications

Fourier and inverse Fourier transform of polynomial f: α=±wj and x=t over IR

Laplace and inverse Laplace transform of polynomial f: α=±s and x=t   over IR

Example: this formula can be also used to calculate the following integrals:

∫f(x)cosxdx  and  ∫f(x)sinxdx

As e(ix)=cosx+isinx (Euler formula) so we can rewrite the integral  (1) as follows:

∫f(x)e(-ix) dx= ∫f(x)[cosx-isinx]dx

= ∫f(x)cosxdx-i∫f(x)sinx dx                                                                   (2)

∫f(x)e (-ix) dx= -[(f(x))/i+(f'(x))/i2 +(f''(x))/i3 +⋯+(f(n)  (x))/i(n+1) +⋯]  (cosx-

isinx) As i2=-1 ;i3=-i ;i4=1 which is periodically return to the same order:i,-1,i,+1

so

∫f(x)e-ix dx= -[(f(x))/i+(f'(x))/(-1)+(f(2) (x))/(-i)+(f(3) (x))/1…+(f(n) (x))/i(n+1) +⋯]

(cosx-isinx)  (3)

By comparing the both sides; real parts with real parts and the imaginary parts with the imaginaries, we find:

∫f(x)cosxdx=f(x)sinx+f'(x)cosx-f(2) (x)sinx-f(3)(x)cosx+⋯

∫f(x)cosxdx=[f(x)-f(2) (x)+f(4) (x)- …]  sinx+ [f'(x)-f(3) (x)+f(5)(x)- …]cosx

And similarly we can calculate ∫f(x)sinxdx as follows:

∫f(x)sinxdx=-[f(x)-f(2)(x)+f(4)(x)- …]cosx+ [f'(x)-f(3)(x)+f(5)(x)- …]sinx