Global Journal of Researches in Engineering, I: Numerical Methods, Volume 23 Issue 1

These examples can be summarized in Table 1, where its derivatives and integrals are given for some elementary functions. Table 1: Examples of calculation of derivatives and integrals y <- 1> y <- 0.5> y <0> y <0.5> y <1.5> SL (x, k) Γ (n+2) Γ (n+1) x n+1 Γ (n+1,5) Γ (n+1) x n+0,5 x n Γ (n+0,5) Γ (n+1) x n-0,5 Γ (n-0,5) Γ (n+1) x n-1,5 Γ (n+1- k ) Γ (n+1) x n- k x 3 /3 0,601x 2.5 x 2 1,504x 1.5 2,256x 0.5 Γ (3- k ) 2 x 2- k e x e x e x e x e x e x sin(x-π/2) sin(x-0,5 · π/2) sin(x) sin(x+0,5 · π/2) sin(x+1,5 · π/2) sin (x+ k · π/2) Differential functions can be a function of 2 or more arguments, for example, SL (x, y, k), where (x) and (y) are two arguments of the same function: SL (x, y, k x , k y ) = 2 · k y + (x – y) · k x , and k x and k y - are still a parameter. In addition, any continuous elementary function can be used as a parameter, including the same differential integral function, for example: ( , , 1, 2) ≔ sin � ∙ 1+ 2 2� (5) Of particular interest is the differential integral function, in which the parameter k is a complex number s, s = a + i · b, although in general, the parameter k can be any function of a real or complex argument. III. R esearch R esults To obtain the differential integral function, we recall the Laplace integral transformation and Borel's theorem. The integral Laplace transform has the form [ ( )] = ( ) = ∫ ( ) − ≡ [ ( ) ∙ − ] <−1> 0< < ∞ ∞ 0 (6) where s = a + i * b is a complex quantity. Here f (t) is the original function, and F(s) is its Laplace image. This is a direct conversion of the original into an image. The inverse Laplace transform (7) ( ) ≔ 1 2 ∙ � ( ) ≡ [ ∙ ( ) ∙ ] <−1> − ∞ < + ∞ + ∙ ∞ − ∙ ∞ it is necessary to find the original of the function by its image. Let's consider one of the main properties of this transformation – the differentiation of the original function. Let L[f(t)] = F (s) . Let's find L[f(t )<1> ] , where f(t) <1> - is the 1st derivative, and L[f (t) <1> ] - is its image. [ ( ) <1> ] = [ ( ) <1> ∙ − ] <−1> 0< < ∞ = − ∙ ( ) 0 < < ∞ + ∙ [ ( ) ∙ ] <−1> 0 < < ∞ (8) L [f (t) <1> ] = s * F (s) – f (0) (9) For f (0) = 0 L [f (t) <1> ] = s * F(s) (10) and the differentiation of the original function corresponds to the multiplication of the image of the function by s. Let's consider another important property – the integration of the original. If g(t) = [f( τ )d τ ] <-1> 0< τ <t , then under zero initial conditions g (t) <1> = f (t) and L[g(t) <1> ] = L[f(t)] = s * L[g(t)] = s * L[[f( τ ) d τ ] <-1> 0 < τ < t] (11) Since L[f(t)] = F (s), then L [[f ( τ )* d τ ] <-1> 0< τ <t = F(s)/s (12) that is, the integration of the function corresponds to the division of the image F (s) by s . If for t → ∞ the function f (t) increases no faster than M * e at , then e -st * f (t) → 0 for t → ∞ and is equal to f (0 ), and Global Journal of Researches in Engineering Volume XxXIII Issue I Version I 12 Year 2023 © 2023 Global Journals ( ) I Application of Differentialintegral Functions