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Question Number 137717 by mnjuly1970 last updated on 05/Apr/21

Commented by Dwaipayan Shikari last updated on 05/Apr/21

$$f\left(a\right)=\frac{1}{2}\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{3-a}{2}\right)=\frac{1}{2}\left(\frac{1-a}{2}\right)\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{1-a}{2}\right)=\frac{1-a}{4}.\frac{\pi }{sin(\frac{\pi }{2}+\frac{\pi }{2}a)}$$$$=\frac{\pi }{4}(1-a)sec\left(\frac{\pi }{2}a\right)$$$$-\frac{1}{3}{f}^{\prime }\left(0\right)=\frac{\pi }{12}$$

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