New Trends In Laplace Type Integral Transforms With Applications

Introduction In this work, we consider some methods consisting Laplace, Stieljes, Fourier transforms to evaluate some certain integrals and series and also to find the solution of some integral equations. The authors have already studied several methods to evaluate series, integrals and solve fractional differential equations, specially the popular Laplace transform method, [1], [2], [3], [4], [5], [6], [7], [8] and this work is a completion for their previous researches. 2000 Mathematics Subject Classification: 26A33; 34A08; 34K37; 35R11.


Introduction
In this work, we consider some methods consisting Laplace, Stieljes, Fourier transforms to evaluate some certain integrals and series and also to find the solution of some integral equations. The authors have already studied several methods to evaluate series, integrals and solve fractional differential equations, specially the popular Laplace transform method, [1], [2], [3], [4], [5], [6], [7], [8] and this work is a completion for their previous researches. Proof: See [14]. ✷ Definition 1.3. Kratzel function was first introduced by Kratzel and then studied by Puri [18] Z ν ρ (z) = ∞ 0 t ν−1 exp(−t ρ − z t )dt, ν ∈ C, ρ ∈ R, for example we have which is in the form of an integral representation of modified Bessel function and can be expressed in terms of Airy function as below. Letν = − 1 3 , x = 2 3 η Example 1.4. Prove that the following relationship holds true Proof. Taking Laplace transform of the Kratzel function we have now changing the order of integrals in suffices to let s = 0in the last integral and making a change of variable t ρ = w to get Lemma 1.5. The following identities hold true Proof: 1 -We use integral representation for modified Bessel function of order v, to get Changing the order of integration to get In the inner integral, introducing the change of variable z After calculating each integral by definition of Laplace transform, one has 2 -In the above relation, let us differentiate with respect to y , after simplifying , one gets +∞ 0 ).

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A. Aghili * and H. Zeinali Then by using inverse of Laplace transform we get now making a change of variable s + λ = p the following result will be obtained At this point, we evaluate the complex integral by virtue of Titchmarsh theorem [3] multiplying the numerator and denominator by the conjugate of the denominator , we ge ln r(r b cos aπ − r a cos bπ) − π(r b sin aπ − r a sin bπ) r a+b (ln 2 r + π 2 ) e −tr dr.
Lemma 1.7. The following relationship holds true Proof. By using Buschman's theorem (see [14]) for k(t) = 1, g(t) = t 3 and therefore on the other hand by using table of Laplace transform we can see that Using the fact that and therefore Corollary 1.8. The following relationship holds true Solution. Let us consider the function f (t) = e −t 2 so we have Making a change of variable t + s 2 = wthe error function will be appeared as below Now using the previous lemma we can write is named Joukowsky map. This function is a conformal mapping (analytic and angle preserving). Jokouwsky map is extensively used in aerodynamics and physics which transforms a circle into an ellipse, because if z = x + iybe a point on a circle of radius rthen and consequently which is the equation of the desired ellipse. One can deduce from the above relationship that if the radius of the circle tends to bthen the ellipse will tend to a line segment on xaxis between the points x = 2aband x = −2ab.
Proof: By using inverse of Laplace transform we have ds.
It is clear that the function F (s) has branch points at s = −λ± λ 2 − µ 2 therefore according to the figure 2 consider the branch cuts. Case 1: Assume that λ 2 −µ 2 > 0 then we integrate the function e st F (s)on the path Øč 1 and then let it tend to infinity. By integrating along the path indicated in the figure 1 and using residue theorem we have New Trends In Laplace Type Integral Transforms 179 Figure 1: One can show that summation of integrals along AB and DC is zero. Then On the other hand if R → ∞one can show that integrals along the arcs C R and C ′ R also tend to zero. Hence we have ds, now we make a change of variables s + λ = pin the right hand side integral to get where R(p) = ( p 2 + (µ 2 − λ 2 ) + p) ν and Ω 1 is obtained by shifting the ellipse Ω in direction of horizontal axis by λ( if λ > 0shifting to the left and if λ < 0 shifting to the right). Therefore it suffices to evaluate the integralI. This integral can be rewritten as follows e (t+a)p dp.
Now let us make a change of variables w = p 2 − (λ 2 − µ 2 ) + p in the above integral, this is in fact the inverse of the Jokouwsky map p = 1 2 ( w + (λ 2 −µ 2 ) w ) so we can transform the ellipse Ω 1 to a circle of radius r. Therefore we have now we make a change of variables z = t−a t+a w to get according to the definition of Jokouwsky map if the ellipse Ω 1 tends to a line segment then r tends to λ 2 − µ 2 therefore one can rewrite the above equation as below On the other hand, using the fact that I ν (x) = 1 2π π −π e x cos α cos να dα (see [15]) then one gets Case 2: Assume that λ 2 − µ 2 < 0 then the branch points of the function F (s) are s = −λ ± i µ 2 − λ 2 therefore we integrate the function e st F (s) along the path Γ 2 then we let Rto tend to infinity. Similarly to case 1 we obtain the following result The following relationship holds true Proof: Let us assume that g(t) = t 0 f (u)du then we know that G(p) = L{g(t); p} = F (p) p in which F (p) = L{f (t); p} so by substituting p = s 2 + 2µs + λ 2 we can write changing the order of integrals we have now we take inverse of Laplace transform Once again change the order of integrals to get Q(s) s 2 + 2µs + λ 2 e st ds du, therefore by using the solution to the previous theorem we obtain

Evaluation of certain series
Lemma 3.1. The following relationships hold true in which F (s) = L{f (t); t → s}.
Proof: By using the definition of Laplace transform the values of the function F (s)for natural arguments will be at this point, we sum the above relationship over all positive integers to get We know that F (s) = 2 (s− 1 2 ) 2 +1 is the Laplace transform of the functionf (t). Therefore substituting in (??) we get the following result which by manipulating can be rewritten as below   Proof: See [19]. ✷ Proof. It is well known that the generating function of Legendre polynomials is as below substituting t = 1 − 1 p in the above relationship we obtain one can rewrite the above relationship as below , on the other hand we know that L{L n (t); p} = 1 p (1 − 1 p ) n and , therefore, if we take inverse Laplace transform of both sides of the above equation, we get Hence we can write If we integrate both sides of the above equation with respect to the variable x from -1 to 1, because of the orthogonality of the Legendre functions, we get the following result Making a change of variable u = 1+x Special case: Let t = ξ = 0in the previous lemma then regarding L n (0) = J 0 (0) = 1 we will have while its inverse is given by where F (p, q) is analytic in the regions Rep > c, Req > c ′ ( [9], [10], [11] , [16]).
Lemma 5.2. The following relationship holds true Proof: By series expansion of Kelvin function of order zero we have now we take Laplace transform of the above relationship with respect to the variables x, y

Using the elementary relation
on the other hand we have also the following expansion for Bessel's function Hence one has and finally we get the following result f (t) (t + y) ρ dt, |argy| < π and its inverse is as below [21] In the special case of ρ=1, the above relationship leads to the ordinary Stieltjes transform on the other hand Stieltjes transform is the second iterate of the Laplace transform hence therefore the final result will be making a change of variable u = w 2 Lemma 6.3. Assume that S{f (x); x → s} = F (s) then the following relationship holds true Proof: By definition of inverse Stieltjes transform we have making a change of variable F (sw) = η we will have in which C is a closed simple path avoiding the origin and a branch cut on the negative x-axis. Therefore we obtain In special case, for q(s) = √ s one has The following relationship holds true Proof: Taking inverse Stieltjes transform we have which could be rewritten as below ✷ Theorem 6.5 (Schouten-Vanderpol for Stieltjes transform ). Consider the function f (t) and its Stieltjes transform F (s) which are analytic over the region Res > s 0 . If q(s) is another analytic function over Res > s 0 . Then the inverse Stieltjes transform of the function F (q(s)) will be obtained as below Proof: Assume that G(s) = F (q(s)) then from the previous lemma we have Example 6.6. Solve the following Stieltjes type singular integral equation Solution. The above integral equation is indeed the definition of Stieltjes transform, therefore by using the inverse of Stieltjes transform and Schouten-Vanderpol theorem we can write Special case: Let us consider the following integral equation Definition 6.7. Elliptic integrals were first investigated by Giulio Fagnano and Leonhard Euler. Generally any function f which could be expressed as below is called an elliptic function in which R is a rational function of its two arguments, P is a polynomial of degree 3 or 4 with no repeated roots, and c is a constant. In general, integrals in this form cannot be expressed in terms of elementary functions. Exceptions to this general rule are when P has repeated roots, or when R(x, y) contains no odd powers of y. However, with the appropriate reduction formula, every elliptic integral can be brought into a form that involves integrals over rational functions and the three Legendre canonical forms (i.e. the elliptic integrals of the first, second and third kind).But complete elliptic functions of the first and second kind can be written as follows Example 6.8. The following relationship holds true Solution. Considering the definition of the generalized Stieltjes transform and changing the variables t + s = w we have which can be rewritten as follows We can rewrite the above relationship by using hyper-geometric functions

Airy functions
George Biddell Airy (1801-1892) was particularly involved in optics for this reason, he was also interested in the calculation of light intensity in the neighborhood of a caustic (see [12,13]). For this purpose, he introduced the function defined by the integral which is the solution of the following differential equation In 1928 Jeffreys introduced the notation used nowadays The KdV equations are attracting many researchers, and a great deal of works has already been done in some of these equations. In this section, we will implement the joint Laplace -Fourier transforms to construct exact solution for a variant of the KdV equation. u t + αu + βu x + γu xxx = Ai(x), u(x, 0) = f (x).
Solution. By taking joint Laplace -Fourier transform of equation and using boundary condition, we get the following transformed equation For the sake of simplicity, let us assume thatτ = iγw 3 − iwβ − α, and using inverse Laplace transform of transformed equation to obtain ; s− > t} = F (w)e τ t + G(w) The inner integrals can be evaluated by convolution for Fourier transform as below Note. Where * denotes convolution for Fourier transform.

Conclusion
The paper is devoted to study Laplace, Stieltjes integral transforms and their applications in evaluating integrals and series. The authors also discussed Laguerre series as well. The one dimensional Laplace and Fourier Transforms provide powerful method for analyzing linear systems. The main purpose of this work is to develop methods for evaluating some special integrals, series and solution to a variant of non-homogenous KdV equation.