n-an-integer-0-1-1-t-n-1-dt-

Question Number 134931 by metamorfose last updated on 08/Mar/21

$${n}\:{an}\:{integer}\:,\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{t}^{{n}} +\mathrm{1}}{dt}=…??\: \\ $$

Answered by Dwaipayan Shikari last updated on 08/Mar/21

∫_0 ^1 (1/(t^n +1))dt =∫_0 ^1 ((1−t^n )/(1−t^(2n) ))dt t^(2n) =u =(1/(2n))∫_0 ^1 ((t^(1−2n) −t^(1−n) )/(1−u))du =(1/(2n))∫_0 ^1 ((u^((1/(2n))−1) −u^((1/(2n))−(1/2)) )/(1−u))du =(1/(2n))ψ((1/(2n))+(1/2))−(1/(2n))ψ((1/(2n)))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}^{{n}} +\mathrm{1}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{{n}} }{\mathrm{1}−{t}^{\mathrm{2}{n}} }{dt}\:\:\:\:\:{t}^{\mathrm{2}{n}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{1}−\mathrm{2}{n}} −{t}^{\mathrm{1}−{n}} }{\mathrm{1}−{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} −{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\psi\left(\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}{n}}\psi\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$

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