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Question Number 63661 by mathmax by abdo last updated on 06/Jul/19

$$let0<a<1findthevalueof{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t^2}dt$$

Commented bymathmax by abdo last updated on 07/Jul/19

$$changement{t}^2=xgive{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t^2}dt={\int }_0^{\mathrm{\infty }}\frac{{\left(x^{\frac{1}{2}}\right)}^{a-1}}{1+x}\frac{1}{2}{x}^{-\frac{1}{2}}dx$$ $$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{a-1}{2}-\frac{1}{2}}}{1+x}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{a}{2}-1}}{1+x}dx=\frac{1}{2}\frac{\pi }{sin\left(\frac{\pi a}{2}\right)}\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t^2}dt=\frac{\pi }{2sin\left(\frac{\pi a}{2}\right)}.$$

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