I have to prove $$ \operatorname{U}\left(a,\frac{1}{2};x^2\right)=\frac{\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(a+\frac{1}{2}\right)\Gamma\left(a\right)}\sum_{n=0}^{\infty}{\frac{\Gamma\left(a+\frac{n}{2}\right)}{n!}(-2x)^n} $$ My Attempt: $$ \operatorname{U}\left(a,\frac{1}{2};x^2\right)=\frac{1}{\Gamma(a)}\int_0^\infty{e^{-x^2t}t^{a-1}(1+t)^{-a-\frac{1}{2}}dt}\\ =\frac{1}{\Gamma(a)}\int_0^\infty{\left(\sum_{n=0}^\infty{\frac{(-x^2t)^n}{n!}}\right)t^{a-1}(1+t)^{-a-\frac{1}{2}}dt}\\ =\frac{1}{\Gamma(a)}\sum_{n=0}^\infty{\frac{(-x^2)^n}{n!}\int_0^\infty {t^{a+n-1}\left(\frac{1}{1+t}\right)^{a+\frac{1}{2}}dt}}\\ =\frac{1}{\Gamma(a)}\sum_{n=0}^\infty{\frac{(-x^2)^n}{n!}\frac{\Gamma(a+n)\Gamma\left(\frac{1}{2}-n\right)}{\Gamma\left(a+\frac{1}{2}\right)}}\\ =\frac{1}{\Gamma(a)\Gamma\left(a+\frac{1}{2}\right)}\sum_{n=0}^\infty{\frac{\Gamma(a+n)\Gamma\left(\frac{1}{2}-n\right)}{n!}(-x^2)^n} $$ How do I proceed further, please help. Thank you.
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Sorry, what's the function U? – Pedro Ignacio Martinez Bruera Jul 10 '21 at 07:10
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@PedroIgnacioMartinezBruera. Kummer hypergeometric function $, _1F_1\left(a;\frac{1}{2};x^2\right)$ – Claude Leibovici Jul 10 '21 at 07:12
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@PedroIgnacioMartinezBruera Kummer's function of second kind – Hisoka Jul 10 '21 at 07:19
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Please edit the title so it gives the reader some idea of what kind of question this is. – Gerry Myerson Jul 10 '21 at 09:14