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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

$${how}\:{can}\:{it}\:{solve}\:{this} \\ $$

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) ))−..)

$$\int_{\mathrm{0}} ^{\infty} {x}^{{n}} \frac{{e}^{{ax}} −{e}^{−{ax}} }{{e}^{{bx}} +{e}^{−{bx}} }{dx} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{\left({a}+{b}\right){x}} {e}^{−\mathrm{2}{bkx}} −{e}^{\left({b}−{a}\right){x}} {e}^{−\mathrm{2}{bkx}} {dx}\: \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \Gamma\left({n}+\mathrm{1}\right)}{\left(\mathrm{2}{bk}−{a}−{b}\right)^{{n}+\mathrm{1}} }−\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \Gamma\left({n}+\mathrm{1}\right)}{\left({a}−{b}+\mathrm{2}{bk}\right)^{{n}+\mathrm{1}} } \\ $$$$={n}!\left(\left(\frac{\mathrm{1}}{\left({b}−{a}\right)^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{3}{b}−{a}\right)^{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{5}{b}−{a}\right)^{{n}+\mathrm{1}} }+...\right)−\frac{\mathrm{1}}{\left({b}+{a}\right)^{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{3}{b}+{a}\right)^{{n}+\mathrm{1}} }−..\right) \\ $$

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