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0-1-x-x-dx-




Question Number 96990 by Ar Brandon last updated on 06/Jun/20
∫_0 ^1 ((√x))^(√x) dx
01(x)xdx
Answered by Sourav mridha last updated on 06/Jun/20
∫_0 ^1 e^(((√x)/2).ln(x)) dx=Σ_(n=0) ^∞ (1/2^n )∫_0 ^∞ ((x^(n/2) .[ln(x)]^n )/(n!))dx      now let ln(x)=−k so we get  =Σ_(n=0) ^∞ (((−1)^n )/(2^n n!))∫_0 ^∞ e^(−((n/2)+1)k) .k^n dk  =Σ_(n=0) ^∞ (((−1)^n )/(2^n 𝚪(n+1))).((𝚪(n+1))/(((n/2)+1)^(n+1) )).[using Laplace transform  =2Σ_(n=0) ^∞ (((−1)^n )/((n+2)^(n+1) )).
01ex2.ln(x)dx=n=012n0xn2.[ln(x)]nn!dxnowletln(x)=ksoweget=n=0(1)n2nn!0e(n2+1)k.kndk=n=0(1)n2nΓ(n+1).Γ(n+1)(n2+1)n+1.[usingLaplacetransform=2n=0(1)n(n+2)n+1.
Commented by Ar Brandon last updated on 12/Jun/20
��thanks
Commented by smridha last updated on 12/Jun/20
welcome
welcome

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