CLAMFIM Bologna Modell 1 @ Clamfm Integral Multpl 8 ottobre 215 professor Danele Rtell danele.rtell@unbo.t 1/2?
Example Evaluate the measure of A = { (x, y) R 2 x 2 + y 2 1 } { (x, y) R 2 y x } 3 y 2/2?
3/2?
1..5 1..5.5 1..5 1. 1/ 2 ( x x/ 3 ) dy dx + 1 1/ 2 ( 1 x 2 x/ 3 dy ) dx = π 8 1 2 3 4/2?
For exercse. The only dffcult pont s 1 x2 dx = 1 ( x ) 1 x 2 2 + arcsn x 5/2?
Teorema (Euler) n=1 1 n = π2 2 6 6/2?
Teorema (Euler) Dmostrazone. n=1 1 n = π2 2 6 Se A = [, ) [, ) consderamo dxdy (1 + y)(1 + x 2 y) A 6/2?
Integramo prma rspetto ad x e po rspetto ad y ottenendo dxdy ( 1 ) (1 + y)(1 + x 2 y) = dx dy 1 + y 1 + x 2 y A 7/2?
Integramo prma rspetto ad x e po rspetto ad y ottenendo dxdy ( 1 ) A (1 + y)(1 + x 2 y) = dx dy 1 + y 1 + x 2 y ( [ ] 1 arctan( y x) x= ) = dy 1 + y y x= 7/2?
Integramo prma rspetto ad x e po rspetto ad y ottenendo dxdy ( 1 ) A (1 + y)(1 + x 2 y) = dx dy 1 + y 1 + x 2 y ( [ ] 1 arctan( y x) x= ) = dy 1 + y y = π 2 dy y(1 + y) = π 2 x= 2u π2 du = u(1 + u 2 ) 2 7/2?
Invertamo l ordne d ntegrazone: dxdy (1 + y)(1 + x 2 y) = A ( ) dy dx (1 + y)(1 + x 2 y) 8/2?
Invertamo l ordne d ntegrazone: dxdy ( (1 + y)(1 + x 2 y) = A = 1 1 x 2 ) dy dx (1 + y)(1 + x 2 y) ( ( 1 1 + y x2 1 + x 2 y ) ) dy dx 8/2?
Invertamo l ordne d ntegrazone: dxdy ( (1 + y)(1 + x 2 y) = A = = 1 1 x 2 1 1 x 2 ) dy dx (1 + y)(1 + x 2 y) ( ( ) 1 1 + y x2 1 + x 2 y [ ln 1 + y ] y= dx 1 + x 2 y y= ) dy dx 8/2?
Invertamo l ordne d ntegrazone: dxdy ( (1 + y)(1 + x 2 y) = A = = = 1 1 x 2 1 1 x 2 ) dy dx (1 + y)(1 + x 2 y) ( ( ) 1 1 + y x2 1 + x 2 y [ ln 1 + y ] y= dx 1 + x 2 y 1 1 x 2 ln 1 x 2 dx y= ) dy dx 8/2?
Invertamo l ordne d ntegrazone: dxdy ( (1 + y)(1 + x 2 y) = A = = = = 1 1 x 2 1 1 x 2 ) dy dx (1 + y)(1 + x 2 y) ( ( ) 1 1 + y x2 1 + x 2 y [ ln 1 + y ] y= dx 1 + x 2 y 1 1 x 2 ln 1 x 2 dx ln x 2 x 2 1 dx = y= ) dy dx 8/2?
Invertamo l ordne d ntegrazone: dxdy ( (1 + y)(1 + x 2 y) = A = = = = = 2 1 1 x 2 1 1 x 2 ) dy dx (1 + y)(1 + x 2 y) ( ( ) 1 1 + y x2 1 + x 2 y [ ln 1 + y ] y= dx 1 + x 2 y 1 1 x 2 ln 1 x 2 dx ln x 2 x 2 1 dx = ln x x 2 1 dx y= ) dy dx 8/2?
Uguaglando ottenamo ln x π2 dx = x 2 1 4. (a) 9/2?
Uguaglando ottenamo ln x π2 dx = x 2 1 4. (a) Spezzamo l domno d ntegraton n (a) far [, 1] e far [1, ) e cambamo la varable x = 1/u nel secondo ntegrale n modo che ln x x 2 1 dx = = 1 1 ln x x 2 1 dx + ln x x 2 1 dx + 1 1 ln x x 2 1 dx ln u u 2 1 du. (b) 9/2?
Da (a) e (b) ottenamo 1 ln x π2 dx = x 2 1 8. (c) 1/2?
Da (a) e (b) ottenamo Ma 1 1 ln x π2 dx = x 2 1 8. (c) + ln x x 2 1 dx = n= 1 (2n + 1) 2 (d) 1/2?
svluppando l denomnatore dell ntegrando a prmo membro d (c) n sere geometrca e usando l teorema d Beppo Lev ottenamo: 1 ln x x 2 1 dx = 1 + ln x 1 x 2dx = n= 1 ( x 2n ln x) dx. (1) 11/2?
svluppando l denomnatore dell ntegrando a prmo membro d (c) n sere geometrca e usando l teorema d Beppo Lev ottenamo: 1 ln x x 2 1 dx = Integrando per part 1 ( x 2n ln x) dx = 1 + ln x 1 x 2dx = n= ] 1 1 [ x2n+1 2n + 1 ln x + 1 ( x 2n ln x) dx. (1) x 2n 2n + 1 dx = 1 (2n + 1) 2 (2) 11/2?
svluppando l denomnatore dell ntegrando a prmo membro d (c) n sere geometrca e usando l teorema d Beppo Lev ottenamo: 1 ln x x 2 1 dx = Integrando per part 1 ( x 2n ln x) dx = 1 + ln x 1 x 2dx = n= ] 1 1 [ x2n+1 2n + 1 ln x + Qund confrontando (c) e (d) ottenamo 1 ( x 2n ln x) dx. (1) x 2n 2n + 1 dx = 1 (2n + 1) (2) 2 + n= 1 (2n + 1) = π2 2 8 11/2?
Danele Rtell: Another proof of ζ(2) = π2 usng double ntegrals. 6 Amercan Mathematcal Monthly: Volume 12 (213) 642 645 https://dl.dropboxusercontent.com/u/1752831/amer.math.mont 12.7.642.pdf 12/2?
Euler functons Euler Gamma, Γ(z) s defned for z > by: Γ(z) = t z 1 e t dt 13/2?
Euler functons Euler Gamma, Γ(z) s defned for z > by: Γ(z) = t z 1 e t It s an mproper ntegral, but t converges dt 13/2?
Euler functons Euler Gamma, Γ(z) s defned for z > by: Γ(z) = t z 1 e t It s an mproper ntegral, but t converges Gamma extends the factoral snce dt Γ(z + 1) = z Γ(z) and beng Γ(1) = 1 f n N we have n! = Γ(n + 1) 13/2?
Γ ( ) 1 2 = + t 1 2 1 e t dt = e t t dt 14/2?
Γ t = u = t = u 2 ( ) 1 2 = + = dt = 2udu t 1 2 1 e t dt = e t t dt 14/2?
Γ t = u = t = u 2 ( ) 1 2 = + = dt = 2udu ( ) 1 Γ = 2 t 1 2 1 e t dt = 2ue u2 u du e t t dt 14/2?
( ) 1 + Γ = t 1 2 1 e t dt = 2 t = u = t = u 2 = dt = 2udu ( ) 1 2ue u2 Γ = du = 2 2 u e t t dt e u2 du 14/2?
Γ t = u = t = u 2 Γ ( ) 1 2 ( ) 1 2 = = + = dt = 2udu 2ue u2 u t 1 2 1 e t dt = du = 2 e t t dt e u2 du = π 14/2?
Una curostà su Ramanujan e la funzone Gamma ( 1) nγ3 (n + 1 2 ) (4n + 1) = π 2(n!) 3 n= 15/2?
Euler Reflexon Formula for x / Z Γ(x)Γ(1 x) = π sn πx 16/2?
Euler Beta Theorem If x, y > : Γ(x) Γ(y) Γ(x + y) = 1 s x 1 (1 s) y 1 ds 17/2?
Euler Beta Theorem If x, y > : Γ(x) Γ(y) Γ(x + y) = 1 s x 1 (1 s) y 1 ds Ths dentty can be reformulated ntroducng the Euler Beta functon: n such a way B(x, y) = 1 s x 1 (1 s) y 1 ds Γ(x) Γ(y) B(x, y) = Γ(x + y) 17/2?
roof We start from Gamma s defnton Γ(x) = varable puttng t = u 2 so that Γ(x) = 2 + + u 2x 1 e u2 du t x 1 e t dt then change 18/2?
roof We start from Gamma s defnton Γ(x) = varable puttng t = u 2 so that Γ(x) = 2 Smlarly Γ(y) = 2 + + + u 2x 1 e u2 du v 2y 1 e v2 dv t x 1 e t dt then change 18/2?
roof We start from Gamma s defnton Γ(x) = varable puttng t = u 2 so that Γ(x) = 2 Smlarly Γ(y) = 2 Now use Fubn s Theorem Γ(x)Γ(y) = 4 + + [,+ ) [,+ ) + u 2x 1 e u2 du v 2y 1 e v2 dv t x 1 e t dt then change u 2x 1 v 2y 1 e (u2 +v 2) dudv 18/2?
Change to polar coordnate u = ρ cos ϑ v = ρ sn ϑ obtanng 19/2?
Change to polar coordnate u = ρ cos ϑ v = ρ sn ϑ obtanng Γ(x)Γ(y) = 4 ( + ) ( π/2 ρ 2x+2y 1 e ρ2 dρ cos 2x 1 ϑ sn 2y 1 ϑ dϑ ) 19/2?
Change to polar coordnate u = ρ cos ϑ v = ρ sn ϑ obtanng Γ(x)Γ(y) = 4 ( + = Γ(x + y) ) ( π/2 ρ 2x+2y 1 e ρ2 dρ π/2 2 cos 2x 1 ϑ sn 2y 1 ϑ dϑ cos 2x 1 ϑ sn 2y 1 ϑ dϑ ) 19/2?
To end the proof we have to show that B(x, y) = π/2 2 cos 2x 1 ϑ sn 2y 1 ϑ dϑ 2/2?
To end the proof we have to show that B(x, y) = π/2 But comng back to Beta s defnton 2 cos 2x 1 ϑ sn 2y 1 ϑ dϑ B(x, y) = 1 we are done puttng s = cos 2 ϑ s x 1 (1 s) y 1 ds 2/2?