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0-pi-2-x-1-2-e-x-dx-




Question Number 127211 by mohammad17 last updated on 27/Dec/20
∫_0 ^( π^2 ) x^(−(1/2)) e^(−x) dx
0π2x12exdx
Answered by Dwaipayan Shikari last updated on 27/Dec/20
∫_0 ^π^2  x^(−(1/2)) e^(−x) dx            x=u^2 ⇒1=2u(du/dx)  =2∫_0 ^π e^(−u^2 ) du=2∫_0 ^∞ e^(−u^2 ) du−2∫_π ^∞ e^(−u^2 ) du  =(√π)−((2e^(−π^2 ) )/(2π+(1/(π+(2/(2π+(3/(π+...))))))))  Or ∫_0 ^π^2  x^(−(1/2)) e^(−x) dx=∫_0 ^∞ x^(−(1/2)) e^(−x) −∫_π^2  ^∞ x^(−(1/2)) e^(−x^2 ) =Γ((1/2))−Γ((1/2),π^2 )  Incomplete Gamma Γ(a,z)=∫_z ^∞ t^(a−1) e^(−t) dt
0π2x12exdxx=u21=2ududx=20πeu2du=20eu2du2πeu2du=π2eπ22π+1π+22π+3π+Or0π2x12exdx=0x12exπ2x12ex2=Γ(12)Γ(12,π2)IncompleteGammaΓ(a,z)=zta1etdt
Commented by mohammad17 last updated on 27/Dec/20
sir is the value infinte or finite ?
siristhevalueinfinteorfinite?
Commented by Dwaipayan Shikari last updated on 27/Dec/20
Finite, It is a continued fraction  Q127186
Finite,ItisacontinuedfractionQ127186
Answered by mindispower last updated on 28/Dec/20
Σ_(k≥0) ∫^π^2  x^(−(1/2)) .(((−x)^k )/(k!))dx  =Σ_(k≥0) (((−1)^k )/(k!))∫_0 ^π^2  x^(k−(1/2)) dx  =Σ_(k≥0) (((−1)^k )/(k!))[(x^(k+(1/2)) /(k+(1/2)))]_0 ^π =Σ_(k≥0) (((−1)^k π^(2k+1) )/((k+(1/2))k!))  =2π+2πΣ_(k≥1) ((Π_(j=0) ^(k−1) ((1/2)+j))/(Π_(j=0) ^(k−1) ((3/2)+j))).(((−π^2 )^k )/(k!))  =2π(1+Σ_(k≥1) ((((1/2))_k )/(((3/2))_k )).(((−π^2 )^k )/(k!)))  =2π_1 F_1 ((1/2);(3/2);−π^2 )
k0π2x12.(x)kk!dx=k0(1)kk!0π2xk12dx=k0(1)kk![xk+12k+12]0π=k0(1)kπ2k+1(k+12)k!=2π+2πk1k1j=0(12+j)k1j=0(32+j).(π2)kk!=2π(1+k1(12)k(32)k.(π2)kk!)=2π1F1(12;32;π2)

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