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Question Number 122625 by mnjuly1970 last updated on 18/Nov/20
... advanced integral... i: ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx=γ ii: ψ(x)=∫_0 ^( ∞) ((e^(−t) /t) −(e^(−tx) /(1−e^(−t) )))dt solution :{_(2 : ln(n) =^(easy) ∫_0 ^( 1) ((x^(n−1) −1)/(ln(x)))dx (∗∗)) ^(1: H_n = Σ_(k=1) ^n (1/k) =∫_0 ^( 1) ((1−x^n )/(1−x)) dx (∗)) (∗)−(∗∗): H_n −ln(n)=∫_0 ^1 (((1−x^n )/(1−x)) −((x^(n−1) −1)/(ln(x))))dx lim_(n→∞) (x^n )=^(0<x<1) 0 lim_(n→∞) (H_n −ln(n))=∫_0 ^( 1) ((1/(1n(x)))+(1/(1−x)))dx γ= ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx ✓ ............................. ψ(x)=^(easy) −γ+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt ψ(x)=−∫_0 ^( 1) (1/(ln(t)))+(1/(1−t))dt+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt =∫_0 ^( 1) −(1/(ln(t))) +((1−t^(x−1) −1)/(1−t))dt =−∫_0 ^( 1) (1/(ln(t)))+(t^(x−1) /(1−t)) dt=^(t=e^(−y) ) =−∫_∞ ^( 0) ((1/(−y))+(e^(−yx+y) /(1−e^(−y) )))(−e^(−y) )dy =∫_0 ^( ∞) (e^(−y) /y)−(e^(−yx) /(1−e^(−y) ))dy ∵ ψ(x)=∫_0 ^( ∞) ((e^(−y) /y)−(e^(−yx) /(1−e^(−y) )))dy ✓
$$\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{integral}… \\ $$$$\:\:\:\:\:{i}:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{ln}\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx}=\gamma \\ $$$$\:\:\:\:\:\:{ii}:\:\psi\left({x}\right)=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{e}^{−{t}} }{{t}}\:−\frac{{e}^{−{tx}} }{\mathrm{1}−{e}^{−{t}} }\right){dt} \\ $$$$\:\:\:\:{solution}\::\left\{_{\mathrm{2}\::\:{ln}\left({n}\right)\:\overset{{easy}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{n}−\mathrm{1}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:\:\:\left(\ast\ast\right)} ^{\mathrm{1}:\:\:{H}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:{dx}\:\:\left(\ast\right)} \right. \\ $$$$\:\:\left(\ast\right)−\left(\ast\ast\right):\:\:{H}_{{n}} −{ln}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:−\frac{{x}^{{n}−\mathrm{1}} −\mathrm{1}}{{ln}\left({x}\right)}\right){dx} \\ $$$$\:\:\:\:\:{lim}_{{n}\rightarrow\infty} \left({x}^{{n}} \right)\overset{\mathrm{0}<{x}<\mathrm{1}} {=}\mathrm{0} \\ $$$$\:\:{lim}_{{n}\rightarrow\infty} \left({H}_{{n}} −{ln}\left({n}\right)\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}{n}\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx} \\ $$$$\:\:\gamma=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{ln}\left({x}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx}\:\:\checkmark \\ $$$$……………………….. \\ $$$$\:\:\:\:\psi\left({x}\right)\overset{{easy}} {=}−\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\psi\left({x}\right)=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{ln}\left({t}\right)}+\frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{{ln}\left({t}\right)}\:+\frac{\mathrm{1}−{t}^{{x}−\mathrm{1}} −\mathrm{1}}{\mathrm{1}−{t}}{dt}\: \\ $$$$\:\:\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{ln}\left({t}\right)}+\frac{{t}^{{x}−\mathrm{1}} }{\mathrm{1}−{t}}\:{dt}\overset{{t}={e}^{−{y}} } {=} \\ $$$$\:\:\:\:\:=−\int_{\infty} ^{\:\mathrm{0}} \left(\frac{\mathrm{1}}{−{y}}+\frac{{e}^{−{yx}+{y}} }{\mathrm{1}−{e}^{−{y}} }\right)\left(−{e}^{−{y}} \right){dy} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{y}} }{{y}}−\frac{{e}^{−{yx}} }{\mathrm{1}−{e}^{−{y}} }{dy}\: \\ $$$$\because\:\psi\left({x}\right)=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{e}^{−{y}} }{{y}}−\frac{{e}^{−{yx}} }{\mathrm{1}−{e}^{−{y}} }\right){dy}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$ \\ $$

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