Question Number 131858 by mohammad17 last updated on 09/Feb/21
Commented by mohammad17 last updated on 09/Feb/21
$${how}\:{can}\:{it}\:{solve}\:{this} \\ $$
Commented by Dwaipayan Shikari last updated on 09/Feb/21
$$\int_{\mathrm{0}} ^{\infty} {x}^{{n}} \frac{{e}^{{ax}} −{e}^{−{ax}} }{{e}^{{bx}} +{e}^{−{bx}} }{dx} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{\left({a}+{b}\right){x}} {e}^{−\mathrm{2}{bkx}} −{e}^{\left({b}−{a}\right){x}} {e}^{−\mathrm{2}{bkx}} {dx}\: \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \Gamma\left({n}+\mathrm{1}\right)}{\left(\mathrm{2}{bk}−{a}−{b}\right)^{{n}+\mathrm{1}} }−\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \Gamma\left({n}+\mathrm{1}\right)}{\left({a}−{b}+\mathrm{2}{bk}\right)^{{n}+\mathrm{1}} } \\ $$$$={n}!\left(\left(\frac{\mathrm{1}}{\left({b}−{a}\right)^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{3}{b}−{a}\right)^{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{5}{b}−{a}\right)^{{n}+\mathrm{1}} }+…\right)−\frac{\mathrm{1}}{\left({b}+{a}\right)^{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{3}{b}+{a}\right)^{{n}+\mathrm{1}} }−..\right) \\ $$