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Question Number 142488 by rs4089 last updated on 01/Jun/21

$${\int }_0^1\frac{t^{k-1}}{1+t^2}dt$$

Answered by Ar Brandon last updated on 01/Jun/21

$$\mathrm{\Phi }={\int }_0^1\frac{{\mathrm{t}}^{\mathrm{k}-1}}{1+{\mathrm{t}}^2}\mathrm{dt}={\int }_0^1\frac{(1-{\mathrm{t}}^2){\mathrm{t}}^{\mathrm{k}-1}}{1-{\mathrm{t}}^4}\mathrm{dt},\mathrm{t}={\mathrm{u}}^{\frac{1}{4}}$$$$=\frac{1}{4}{\int }_0^1\frac{{\mathrm{u}}^{\frac{\mathrm{k}}{4}-1}-{\mathrm{u}}^{\frac{\mathrm{k}}{4}-\frac{1}{2}}}{1-\mathrm{u}}\mathrm{du}=\frac{1}{4}(\psi (\frac{\mathrm{k}}{4}+\frac{1}{2})-\psi \left(\frac{\mathrm{k}}{4}\right))$$

Answered by Dwaipayan Shikari last updated on 01/Jun/21

$${\int }_0^1\frac{t^{k-1}}{1+t^2}dt$$$$=\frac{1}{4}{\int }_0^1\frac{u^{\frac{k}{4}-1}-u^{\frac{k}{4}+\frac{1}{2}-1}}{1-u}du$$$$=\frac{1}{4}(\psi (\frac{k}{4}+\frac{1}{2})-\psi \left(\frac{k}{4}\right))$$

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