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I am looking for the closed expresion of the following integral

$$I_s=\int_0^1\frac{(1-t^2)^s}{2^{-s}(1-t^2)^s+ t^s}\frac{dt}{1+t^2} $$

I have proved already that $I_s$ exists for any $s\in \Bbb R.$ For $s=0$ I got

$$I_0=\frac{\pi}{8}.$$

Guy Fsone
  • 23,903

1 Answers1

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Using changing variables $t=\tan\frac{x}{2}$ then $dt=\frac{1}{2}(1+\tan^2{\frac{x}{2}})dx=\frac{1}{2}(1+t^2)dx,$ so $$I_s=\int_0^1\frac{(1-t^2)^s}{2^{-s}(1-t^2)^s+ t^s}\frac{dt}{1+t^2} =2^{s-1}\int_0^1\frac{1}{1+(\frac{2t}{1-t^2})^s}\frac{2dt}{(1+t^2)}\\=2^{s-1}\int_0^{\frac\pi2}\frac{1}{1+\tan^{s} x} \,dx =\color{red}{2^{s-3}\pi}$$ the last integral which can be calculated as per the method shown here.