Let $F:Omegatimes(0,infty)rightarrow mathbb{C}$
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Let $F:Omegatimes(0,infty)rightarrow mathbb{C}$ such that $F(z,t)$ is analytic on $Omega$ for each $tin(0,infty)$ and $F$ and $frac{d}{dz}F$ are continuous on $Omegatimes(0,infty$). And for any $Ksubseteq Omega$ compact, there is a Riemann integrable $M_k:(0,infty)rightarrow[0,infty)$ such that $sup_{zin K}|F(z,t)|leq M_K (t)$ and $int_{0}^{infty}M_k(t)dt$
If $$f(z)=int_{0}^{1}F(z,t)dt$$ show that $$f'(z)=int_0^1frac{d}{dz}F(z,t)dt$$ And if $$g(z)=int_a^b F(z,t)dt hbox{ where } 0<a<b$$ prove that $$g'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$ Lastly, if $$f(z)=int_0^{infty}F(z,t)dt$$ then, $$f'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$
complex-analysis
asked Nov 17 at 19:13
Ya G
35019
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Let $F:Omegatimes(0,infty)rightarrow mathbb{C}$ such that $F(z,t)$ is analytic on $Omega$ for each $tin(0,infty)$ and $F$ and $frac{d}{dz}F$ are continuous on $Omegatimes(0,infty$). And for any $Ksubseteq Omega$ compact, there is a Riemann integrable $M_k:(0,infty)rightarrow[0,infty)$ such that $sup_{zin K}|F(z,t)|leq M_K (t)$ and $int_{0}^{infty}M_k(t)dt$
If $$f(z)=int_{0}^{1}F(z,t)dt$$ show that $$f'(z)=int_0^1frac{d}{dz}F(z,t)dt$$ And if $$g(z)=int_a^b F(z,t)dt hbox{ where } 0<a<b$$ prove that $$g'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$ Lastly, if $$f(z)=int_0^{infty}F(z,t)dt$$ then, $$f'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$
complex-analysis
asked Nov 17 at 19:13
Ya G
35019
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $F:Omegatimes(0,infty)rightarrow mathbb{C}$ such that $F(z,t)$ is analytic on $Omega$ for each $tin(0,infty)$ and $F$ and $frac{d}{dz}F$ are continuous on $Omegatimes(0,infty$). And for any $Ksubseteq Omega$ compact, there is a Riemann integrable $M_k:(0,infty)rightarrow[0,infty)$ such that $sup_{zin K}|F(z,t)|leq M_K (t)$ and $int_{0}^{infty}M_k(t)dt$
If $$f(z)=int_{0}^{1}F(z,t)dt$$ show that $$f'(z)=int_0^1frac{d}{dz}F(z,t)dt$$ And if $$g(z)=int_a^b F(z,t)dt hbox{ where } 0<a<b$$ prove that $$g'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$ Lastly, if $$f(z)=int_0^{infty}F(z,t)dt$$ then, $$f'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$
complex-analysis
asked Nov 17 at 19:13
Ya G
35019
Let $F:Omegatimes(0,infty)rightarrow mathbb{C}$ such that $F(z,t)$ is analytic on $Omega$ for each $tin(0,infty)$ and $F$ and $frac{d}{dz}F$ are continuous on $Omegatimes(0,infty$). And for any $Ksubseteq Omega$ compact, there is a Riemann integrable $M_k:(0,infty)rightarrow[0,infty)$ such that $sup_{zin K}|F(z,t)|leq M_K (t)$ and $int_{0}^{infty}M_k(t)dt$
If $$f(z)=int_{0}^{1}F(z,t)dt$$ show that $$f'(z)=int_0^1frac{d}{dz}F(z,t)dt$$ And if $$g(z)=int_a^b F(z,t)dt hbox{ where } 0<a<b$$ prove that $$g'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$ Lastly, if $$f(z)=int_0^{infty}F(z,t)dt$$ then, $$f'(z)=int_0^{infty}frac{d}{dz}F(z,t)dt$$
complex-analysis
complex-analysis
asked Nov 17 at 19:13
Ya G
35019
asked Nov 17 at 19:13
Ya G
35019
asked Nov 17 at 19:13
Ya G
35019
asked Nov 17 at 19:13
Ya G
35019
asked Nov 17 at 19:13
Ya G
35019
35019
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