Question Number 154223 by mnjuly1970 last updated on 15/Sep/21
![∫_0 ^( 1) (x^( i+1) /(1−x)) dx = [−ln(1−x)x^( i+1) ]_0 ^( 1) + (1+i)∫_0 ^( 1) x^( i) ln (1−x )dx = (1+i ) ∫_0 ^( 1) Σ_(n=1) ^∞ (x^( n+i) /n)dx = (1+i )Σ(1/(n (n+i+1 )))= = Σ(1/(n )) −(1/(n+i+1)) = γ + ψ ( i+2 ) = γ + (1/(1+i)) +ψ (1+i) γ + ((1−i)/2) + (1/i) + ψ (i ) im = −(3/2) + (1/2) +(π/2) coth(π ) −1 +(π/2) tan h(π ) ..✓](https://www.tinkutara.com/question/Q154223.png)
$$ \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\:{i}+\mathrm{1}} }{\mathrm{1}−{x}}\:{dx}\:=\:\left[−{ln}\left(\mathrm{1}−{x}\right){x}^{\:{i}+\mathrm{1}} \right]_{\mathrm{0}} ^{\:\mathrm{1}} \\ $$$$\:\:\:+\:\left(\mathrm{1}+{i}\right)\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:{i}} \:{ln}\:\left(\mathrm{1}−{x}\:\right){dx} \\ $$$$\:\:\:=\:\left(\mathrm{1}+{i}\:\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\:{n}+{i}} }{{n}}{dx} \\ $$$$\:=\:\:\left(\mathrm{1}+{i}\:\right)\Sigma\frac{\mathrm{1}}{{n}\:\left({n}+{i}+\mathrm{1}\:\right)}= \\ $$$$\:=\:\Sigma\frac{\mathrm{1}}{{n}\:}\:−\frac{\mathrm{1}}{{n}+{i}+\mathrm{1}}\:=\:\gamma\:+\:\psi\:\left(\:{i}+\mathrm{2}\:\right) \\ $$$$\:\:\:=\:\gamma\:+\:\frac{\mathrm{1}}{\mathrm{1}+{i}}\:+\psi\:\left(\mathrm{1}+{i}\right) \\ $$$$\:\:\:\:\:\gamma\:+\:\frac{\mathrm{1}−{i}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{{i}}\:+\:\psi\:\left({i}\:\right) \\ $$$$\:\:\:{im}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\:{coth}\left(\pi\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:+\frac{\pi}{\mathrm{2}}\:{tan}\:{h}\left(\pi\:\right)\:..\checkmark \\ $$