I-0-a-1-n-x-x-dx-I-

Question Number 6983 by FilupSmith last updated on 04/Aug/16

$${I}=\int_{\mathrm{0}} ^{\:{a}} \left(\mathrm{1}+\frac{{n}}{{x}}\right)^{{x}} {dx} \\ $$$${I}=? \\ $$

Commented by FilupSmith last updated on 04/Aug/16

Current working I=∫_0 ^( a) (1+(n/x))^x dx I=∫_0 ^( a) e^(xln(1+(n/x))) dx e^(xln(1+(n/x))) =Σ_(t=0) ^∞ ((x^t ln(1+(n/x))^t )/(t!)) ∴I=Σ_(t=0) ^∞ ∫_0 ^( a) (((xln(1+(n/x)))^t )/(t!))dx u=ln(1+(n/x)) ⇒ x=(((e^u −1)/n))^(−1) dx=−(((e^u −1)/n))^(−2) (e^u /n)du dx=−(e^u /(n(((e^u −1)/n))^2 ))du dx=−e^u ((((√n)(e^u −1))/n))^(−2) du dx=−e^u (((e^u −1)/( (√n))))^(−2) du dx=−e^u (1/n)(((e^u −1)/1))^(−2) du ∴I=Σ_(t=0) ^∞ (1/(t!))∫_∞ ^( ln(1+(n/a))) (u(((e^u −1)/n))^(−1) )^t (−(e^u /n)((1/(e^u −1)))^2 )du ∴I=Σ_(t=0) ^∞ (−1)(1/(t!))∫_∞ ^( ln(1+(n/a))) u^t ((n/(e^u −1)))^t (e^u /n)((1/(e^u −1)))^2 du ∴I=Σ_(t=0) ^∞ (1/(nt!))∫_(ln(1+(n/a))) ^( ∞) u^t e^u (n^t /((e^u −1)^t )) (1/((e^u −1)^2 ))du ∴I=Σ_(t=0) ^∞ (n^t /(nt!))∫_(ln(1+(n/a))) ^( ∞) ((u^t e^u )/((e^u −1)^(t+2) )) du ∴I=Σ_(t=0) ^∞ (n^(t−1) /(t!))∫_(ln(1+(n/a))) ^( ∞) ((u^t e^u )/((e^u −1)^(t+2) )) du

$$\mathrm{Current}\:\mathrm{working} \\ $$$${I}=\int_{\mathrm{0}} ^{\:{a}} \left(\mathrm{1}+\frac{{n}}{{x}}\right)^{{x}} {dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\:{a}} {e}^{{x}\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{x}}\right)} {dx} \\ $$$${e}^{{x}\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{x}}\right)} =\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{t}} \mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{x}}\right)^{{t}} }{{t}!} \\ $$$$\therefore{I}=\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:{a}} \frac{\left({x}\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{x}}\right)\right)^{{t}} }{{t}!}{dx} \\ $$$${u}=\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{x}}\right)\:\:\Rightarrow\:\:{x}=\left(\frac{{e}^{{u}} −\mathrm{1}}{{n}}\right)^{−\mathrm{1}} \\ $$$${dx}=−\left(\frac{{e}^{{u}} −\mathrm{1}}{{n}}\right)^{−\mathrm{2}} \frac{{e}^{{u}} }{{n}}{du} \\ $$$${dx}=−\frac{{e}^{{u}} }{{n}\left(\frac{{e}^{{u}} −\mathrm{1}}{{n}}\right)^{\mathrm{2}} }{du} \\ $$$${dx}=−{e}^{{u}} \left(\frac{\sqrt{{n}}\left({e}^{{u}} −\mathrm{1}\right)}{{n}}\right)^{−\mathrm{2}} {du} \\ $$$${dx}=−{e}^{{u}} \left(\frac{{e}^{{u}} −\mathrm{1}}{\:\sqrt{{n}}}\right)^{−\mathrm{2}} {du} \\ $$$${dx}=−{e}^{{u}} \frac{\mathrm{1}}{{n}}\left(\frac{{e}^{{u}} −\mathrm{1}}{\mathrm{1}}\right)^{−\mathrm{2}} {du} \\ $$$$\therefore{I}=\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{t}!}\int_{\infty} ^{\:\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{a}}\right)} \left({u}\left(\frac{{e}^{{u}} −\mathrm{1}}{{n}}\right)^{−\mathrm{1}} \right)^{{t}} \left(−\frac{{e}^{{u}} }{{n}}\left(\frac{\mathrm{1}}{{e}^{{u}} −\mathrm{1}}\right)^{\mathrm{2}} \right){du} \\ $$$$\therefore{I}=\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)\frac{\mathrm{1}}{{t}!}\int_{\infty} ^{\:\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{a}}\right)} {u}^{{t}} \left(\frac{{n}}{{e}^{{u}} −\mathrm{1}}\right)^{{t}} \frac{{e}^{{u}} }{{n}}\left(\frac{\mathrm{1}}{{e}^{{u}} −\mathrm{1}}\right)^{\mathrm{2}} {du} \\ $$$$\therefore{I}=\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{nt}!}\int_{\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{a}}\right)} ^{\:\infty} {u}^{{t}} {e}^{{u}} \frac{{n}^{{t}} }{\left({e}^{{u}} −\mathrm{1}\right)^{{t}} }\:\frac{\mathrm{1}}{\left({e}^{{u}} −\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$$\therefore{I}=\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{{t}} }{{nt}!}\int_{\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{a}}\right)} ^{\:\infty} \frac{{u}^{{t}} {e}^{{u}} }{\left({e}^{{u}} −\mathrm{1}\right)^{{t}+\mathrm{2}} }\:{du} \\ $$$$\therefore{I}=\underset{{t}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{{t}−\mathrm{1}} }{{t}!}\int_{\mathrm{ln}\left(\mathrm{1}+\frac{{n}}{{a}}\right)} ^{\:\infty} \frac{{u}^{{t}} {e}^{{u}} }{\left({e}^{{u}} −\mathrm{1}\right)^{{t}+\mathrm{2}} }\:{du} \\ $$

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