Complex Inversion Formula for Exponential Integral Transform with Applications

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 3, 28, no. 6, Complex Inversion Formula for Exponential Integral Transform with Applications A. Aghili and Z. Kavooci Department of Mathematics, Faculty of Sciences University of Guilan, P.O. Box 84, Rasht, Iran Zahra Kavooci Abstract In this article, we derive a complex inversion formula and some new theorems related to Stieltjes-transform and exponential integral transform defined in [2],[3],we also give some illustrative examples. Mathematics Subject Classification: Primary 44A; Secondary 44A35 Keywords: Ξ -transform, Laplace transform, Stieltjes-transform Introduction Integral transforms are used to accomplish the solution of certain problems with less effort and in a simple routine way. Laplace transforms provides a powerful method for solving differential and integral equations. Recently, many other Transforms have been developed, but most have limited applicability.this work also contains some discussion on several other integral transforms that have been used recently successfully in the solution of certain boundary value problems and in other applications. Included in this category are, Stieltjes-transform, exponential integral transform ([],[2],[3])which have also been given.the Laplace-transform of the function is defined as L{f(t)} = F (s) = exp( st)f(t)dt. (-) The inverse laplace-transform is given by the integral relationship σ+i f(t) =L {F (s)} = F (s)e st ds, (-2) 2πi σ i where σ is any suitably chosen large positive constant.this improper integral is a contour integral taken along the vertical line s = σ + iτ in complex

2 782 A. Aghili and Z. Kavooci s-plane. for more detail see[4]. Example. : The following are immediate consequences of definition (-). L{δ(x a)} = exp( as). 2. L{exp( ax)} = s + a. 3. L{ x + a } = e as E (as). Example 2.:Using complex inversion formula for Laplace transform to find s + f(t) =L { }. s Solution : We have the following relation, to s + = s s + + s s +, therefore, it is sufficient to evaluate L { s+ }, using relation (-2), leads f(t) = c+i e st ds, 2πi c i s + at this point, in order to avoid the complex integration around a complicated key-hole contour, we use the integral representation for s+ as follows, 2 + e (s+)u2 du = π s +. Inserting in the above relationship, we get, f(t) = c+i { 2 e (s+)u2 du} e st ds, 2πi c i π changing the order of integration, to obtain f(t) = 2 e u2 du{ c+i e (t u2)s ds}, π 2πi c i but, the value of inner integral, is δ(t u 2 ) therefore, f(t) = e u2 δ(t u 2 )du, π

3 Complex inversion formula 783 letting u 2 = x and using some elementary properties of delta function, one has finally L { } = f(t) = e t, s + πt and, hence, finally, s + L { } = L { } + L { s s + s s + }, s + L { } = e t t + H(t x) e x dx, s πt πx s + L { } = e t + erf( t). s πt 2 Stieltjes Transform Stieltjes Transform is defined as follows f(t)dt {f(t),y} = t + y. It is well-known that the second iterate of the Laplace transform is the Stieltjestransform [2],[3], that is, L 2 {f(t); y} = L{L{f(t); u},y} = {f(t),y} = F (y). (2-) Example 2.: For f(t) =δ(t a), we can verify the above relationship we have δ(t a)dt {δ(t a),y} = = t + y a + y, on the other hand, second iterate of the Laplace transform of δ(t a) is, L{δ(t a),u} = e au, and L{e au,y} = y + a. Example 2.2: show that Q.E.D {t μ,y} = πyμ sin μπ ( <μ<). Solution :Upon using relationship(3-)

4 784 A. Aghili and Z. Kavooci L 2 t μ {f(t); y} = L{L{f(t); u},y} = {f(t),y} = t + y dt = L 2 {f(t); y} = L{L{f(t); u},y} = L{L{t μ,u),y} = L{ Γ(μ) u,y} μ = Γ(μ)L{u μ,y} =Γ(μ)Γ( μ)y μ = π sin μπ yμ ( <μ<). 2. Complex inversion formula for Stieltjes Transform The inverse Stieltjes Transform is given by the following integral relationship f(t) = {F (y)} = λ+i σ+i ( F (y)e yx dy)e xt dx, (2-2) (2πi) 2 λ i σ i where σ,λ are any suitably chosen large positive constants.this improper integral is a contour integral taken along the vertical lines y = σ + iτ and x = λ + iη in complex-plane. Example 2.3: Show that {π 2 y 3 4,t} = t 3 4. Proof : Since Stieltjes transform is second iterate of Laplace transform, we can re-write relation (2-2) as follows finally, f(t) =L {L {π 2 y 3 4,x},t}, using the fact that, t ν = L { Γ(ν+) } hence, s ν+ L {π 2 y 3 π 2 4,x} = Γ( 3) x 4 f(t) =L {L {π 2 y 3 4,x},t} = L { π 2 π 2 3 Γ( 3) x 4,t} = Γ( )Γ( 3) t 4 = t Example 2.4: Evaluate {cos t} and {sin t}. Solution : Let us assume that Γ R be the closed contour composed of a quarter circle z = Re iθ, θ π in first quadrant and two line segments 2 [,R],[,iR] traversed in counter clockwise, and let F (z) = eiz,a>, then z+a we have e iz dz = z + a Γ R 4,

5 Complex inversion formula 785 now, we expand the above integral to obtain R e ix e (iriθ ) Re iθ + a ireiθ dθ + R π x + a dx + 2 taking limit as R tends to infinity, to get e ix x + a dx + π lim 2 R + e i(iy) iy + a e (iriθ ) Re iθ + a ireiθ dθ + + idy =, e i(iy) idy =, iy + a the second integral tends to zero ( in absolute value) therefore, one has the following e ix x + a dx + e i(iy) idy =, + iy + a or, e ix x + a dx = ye y dy + y 2 + a + i 2 ae y y 2 + a 2 dy. Separating real and imaginary parts of both sides of the above relationship, to get cos x + (cos x) = x + a dx = xe x x 2 + a dx, 2 (sin x) = 3 Ξ -Transform E -Transform is defined as follows Ξ {f(t),y} = sin x + x + a dx = ae x x 2 + a 2 dx. exp(xy)e (xy)f(x)dx where E (x) is the exponential integral defined as exp( ξ) E (x) = dξ x ξ It is well-known that the third iterate of the Laplace transform is the Ξ - Transform [2],[3], that is, L 3 {f(t); y} = L{L{L{f(t); u},v},y)=ξ {f(t),y} = F (y). (3-) Example 3. : The following relations hold true. Ξ {δ(x a)} = e as E (as).

6 786 A. Aghili and Z. Kavooci 2. Ξ (x λ,y} = π Γ( λ) sin(λπ) yλ. Theorem 3. : The following identities hold true, k { Ξ {f(x); y} = {L{g(x); s + u}; y} = L{e xy E (xy)g(x); s}. {L{g(x); s + u}; y} = y L{E (x)g( x y ); s y }. { e ax π,y} = πx y exp(ay)erfc( ay). 2 k π x 3 exp( x ),y} = {E (x),y} = L{ 2y kπ exp( k 3 y )y ln(x +) ; y}. x k 2 Erf( y ). e x( y) E (xy)f(x)dx = L{E (xy)f(x); y}. {x exp( ax),y} = a yeay E (ay). Proof : f(x) = e sx g(x) L{f(x); u} = L{e sx g(x); u} = = e x(s+u) g(x)dx = L{g(x); s + u} e sx e ux g(x)dx {L{f(x); u}; y} = {L{g(x); s + u}; y} {L{f(x); u}; y} = L{e sx g(x);u} du u+y = u+y { e ux e sx g(x)dx}du If we change the order of integration and letting y + u = z in the inner integral, we get,

7 Complex inversion formula 787 or, e sx g(x)dx e ux y + u du = e sx e (z y)x g(x)dx y z e zx e sx g(x)dx e yx dz, y z making a new change of variable xz = t, one has e t e sx g(x)dx{ e xy dt} = e sx g(x)e xy E (xy)dx = L{e xy E (xy)g(x); s}. xy t Proof 2 : Setting xy = t, in part of Theorem 3. to get, {L{g(x); s+u}; y} = e s t y E (t)g( t y ) et y dt = y dz, e x( s y ) E (x)g( x y )dx, or finally, = y L{E (x)g( x y ); s y }. Proof 3: { e ax ; y} = L{L{ e ax e zx e ax ; z}; y} = L{ x x 2 2 dx; y} = L{L{ ; z + a}; y} = πx πx π π L{ ; y} = e zy (z + a) (z + a) 2 dz, 2 Let z + a = t2 y to get L{ ; y} = e ay 2 π (z + a) 2 y e t2 dt = e ay ay y erfc( ay). Proof 5: {E (x)} = x + y { e u x u du}dx = e u u u { x + y dx}du = e u u (ln u + y )du y At this point, we set Proof 6: E {f(x); y} = u+y y =x yx ln(x +) ln(x +) e dx = L{ ; y}. x x e x( y) E (xy)f(x)dx = L{E (xy)f(x); y}.

8 788 A. Aghili and Z. Kavooci Proof 7: {x exp( ax),y} = making a change of variable x + y = u a xe ax x + y dx,, one has a yeay E (ay) = a yeay E (ay). 3. Complex inversion formula for Ξ Transform The inverse Ξ Transform is given by the following integral relationship f(t) =Ξ {F (y)} = μ+i λ+i σ+i ( ( F (y)e yx dy)e xz dx)e zt dz (2πi) 3 μ i λ i σ i (3-2) where σ, λ, μ are any suitably chosen large positive constants. This improper integral is a contour integral taken along the vertical lines y = σ + iτ, x = λ + iη and z = μ + iς in complex - plane. Example 3.2 : Show that Ξ Γ( λ {π sin(λπ) yλ ; u} = x λ. Solution: L {L {L πγ( λ) { sin(λπ) yλ ; u}; v}x} = L {L πγ( λ) { sin(λπ)γ( λ) u λ ; v}; x} = L πγ( λ) { sin(λπ)γ( λ)γ(λ) vλ ; x} = πγ( λ) sin(λπ)γ( λ)γ(λ)γ( λ) x λ = x λ. Example 3.3 : Show that L{f(x); y} {g(u); y}dy = f(x) Ξ {g(u); x}dx. Solution: f(x)[ L{f(x); y} {g(u); y}dy = e xy {g(u); y}dy]dx = {g(u); y}[ e xy f(x)dx]dy = f(x)l{ {g(u); y}; x}dx = f(x)ξ {g(u); x}dx.

9 Complex inversion formula 789 Example 3.4 : E {f(x); y} {g(u); y}dy = Solution: E {f(x); y} {g(u); y}dy = g(u)[ Example 3.5 : E {f(x); y} {g(u); y}dy = Solution: g(u) {E {f(x); y}; u}du. g(u) E {f(x); y}[ u + y du}dy = E {f(x); y} dy]du = g(u) {E {f(x); y}; u}du. u + y E {f(x); y} {g(u); y}dy = {f(x); y} E {g(u); y}dy. L{ {f(x); v]; y} {g(u); y}dy = {g(u); y}[ e yv {f(x); v}dv}dy, {f(x); v}[ e yv {g(u); y}dy]dv = {f(x); v}l{ {g(u); y}; v}dv = {f(x); v}e {g(u); v}dv = {f(x); y}e {g(u); y}dy. Example 3.6 : Show that sin λπ L{ πγ(λ) u + a ; y}yλ dy = a λ. Solution:We have from Example 3.5 sin λπ πγ(λ) L{ u + a ; sin λπ y}yλ dy = πγ(λ) = sin2 λπ E π 2 {δ(x a); y} {x λ ; y} Ex5.5 sin2 λπ Γ(λ) π 2 Γ(λ) = sin λπ π y λ sin λπ dy = {y λ ; a} = y + a π x λ L{ {δ(x a); u}; y}p { Γ( λ)γ(λ) ; y}dy {δ(x a); y}e {x λ ; y}dy sin λπ Γ(λ)Γ( λ)a λ = a λ. π

10 79 A. Aghili and Z. Kavooci 4 Conclusion In this article, authors provide complex inversion formula for Stieltjes - transform and Exponential integral transform. Some identities involving these new transforms are given.our primary objective is to introduce methods to use the integral transform rather than concerning ourselves too deeply with the general theorems.we have included a large number of worked examples to illustrate the applicability of these new transforms. We conclude by remarking that many identities involving various integral transforms can be obtained and some other infinite integrals can be evaluated by applying the results considered here. References [] A. Aghili et al: An inversion technique for L 2 -transform with applications, Int. J. Contemp. Math. Sciences, Vol. 2, 27, no. 28, [2] D. Brown et al., Identities for the exponential integral and the complementary error transforms, Applied Mathematics and Computations 82, 26, [3] D. Brown et al., Identities for the E 2, -Transforms and their applications, Applied Mathematics and computations, 82, 26, [4] A. D. Osborn, Complex analysis with applications, Mc.Graw Hill 999. Received: October 26, 27