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Question Number 131858 by mohammad17 last updated on 09/Feb/21

Commented by mohammad17 last updated on 09/Feb/21

how can it solve this

howcanitsolvethis

Commented by Dwaipayan Shikari last updated on 09/Feb/21

∫_0 ^∞ x^n ((e^(ax) −e^(−ax) )/(e^(bx) +e^(−bx) ))dx =Σ_(k=1) ^∞ (−1)^(k+1) ∫_0 ^∞ x^n e^((a+b)x) e^(−2bkx) −e^((b−a)x) e^(−2bkx) dx =Σ_(k=1) ^∞ (((−1)^(k+1) Γ(n+1))/((2bk−a−b)^(n+1) ))−(((−1)^(k+1) Γ(n+1))/((a−b+2bk)^(n+1) )) =n!(((1/((b−a)^(n+1) ))−(1/((3b−a)^(n+1) ))+(1/((5b−a)^(n+1) ))+...)−(1/((b+a)^(n+1) ))+(1/((3b+a)^(n+1) ))−..)

∫0∞xneax−e−axebx+e−bxdx=∑∞k=1(−1)k+1∫0∞xne(a+b)xe−2bkx−e(b−a)xe−2bkxdx=∑∞k=1(−1)k+1Γ(n+1)(2bk−a−b)n+1−(−1)k+1Γ(n+1)(a−b+2bk)n+1=n!((1(b−a)n+1−1(3b−a)n+1+1(5b−a)n+1+...)−1(b+a)n+1+1(3b+a)n+1−..)

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