william_33
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- Joined
- Mar 4, 2013
- Messages
- 10
For each fixed \(\displaystyle \,n\,\), show that \(\displaystyle \displaystyle{\,f_n(z)\,=\,\int_1^{n}\,t^{z-1}\,e^{-t}\,dt\,}\) is a continuous function of \(\displaystyle \,z\,\) and show that \(\displaystyle \displaystyle{\,g(z)\,=\,\int_0^1\,t^{z-1}\,e^{-t}\,dt\,}\) is analytic for \(\displaystyle \,\Re{z}\,>\,0.\)
What I know is that at \(\displaystyle \displaystyle{\,g_n(z)\,=\,\int_{\frac{1}{n}}^1\,t^{z-1}\,e^{-t}\,dt\,}\) I must show that for each \(\displaystyle \,g_n(z),\,n\,\) is analytic and that \(\displaystyle \,g_n\,\) converges uniformly to \(\displaystyle \,g\,\) on closed bounded sets of \(\displaystyle \,\Re z\,>\,0\,\), but I do not know how to prove this?
What I know is that at \(\displaystyle \displaystyle{\,g_n(z)\,=\,\int_{\frac{1}{n}}^1\,t^{z-1}\,e^{-t}\,dt\,}\) I must show that for each \(\displaystyle \,g_n(z),\,n\,\) is analytic and that \(\displaystyle \,g_n\,\) converges uniformly to \(\displaystyle \,g\,\) on closed bounded sets of \(\displaystyle \,\Re z\,>\,0\,\), but I do not know how to prove this?
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