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0-1-t-k-1-1-t-2-dt-

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Question Number 142488 by rs4089 last updated on 01/Jun/21
∫_0 ^1 (t^(k−1) /(1+t^2 ))dt
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{k}−\mathrm{1}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$
Answered by Ar Brandon last updated on 01/Jun/21
Φ=∫_0 ^1 (t^(k−1) /(1+t^2 ))dt=∫_0 ^1 (((1−t^2 )t^(k−1) )/(1−t^4 ))dt, t=u^(1/4) =(1/4)∫_0 ^1 ((u^((k/4)−1) −u^((k/4)−(1/2)) )/(1−u))du=(1/4)(ψ((k/4)+(1/2))−ψ((k/4)))
$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}^{\mathrm{k}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{t}^{\mathrm{k}−\mathrm{1}} }{\mathrm{1}−\mathrm{t}^{\mathrm{4}} }\mathrm{dt},\:\mathrm{t}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\frac{\mathrm{k}}{\mathrm{4}}−\mathrm{1}} −\mathrm{u}^{\frac{\mathrm{k}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−\mathrm{u}}\mathrm{du}=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{k}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{k}}{\mathrm{4}}\right)\right) \\ $$
Answered by Dwaipayan Shikari last updated on 01/Jun/21
∫_0 ^1 (t^(k−1) /(1+t^2 ))dt =(1/4)∫_0 ^1 ((u^((k/4)−1) −u^((k/4)+(1/2)−1) )/(1−u))du =(1/4)(ψ((k/4)+(1/2))−ψ((k/4)))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{k}−\mathrm{1}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\frac{{k}}{\mathrm{4}}−\mathrm{1}} −{u}^{\frac{{k}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}−{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{{k}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{k}}{\mathrm{4}}\right)\right) \\ $$

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