MHB Proving Continuity in a Rectangle Using f(x,y) Function

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If $$f(x,y)$$ be a continuous function of $$(x,y)$$ in the rectangle $$R:{a \leq x \leq b, c \leq y \leq d}$$ , then $$\int_a^b f(x,y) dx$$ is also a continuous function of $$y$$ in $$[c,d]$$

How to proceed with the proof of the above theorem?
 
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suvadip said:
If $$f(x,y)$$ be a continuous function of $$(x,y)$$ in the rectangle $$R:{a \leq x \leq b, c \leq y \leq d}$$ , then $$\int_a^b f(x,y) dx$$ is also a continuous function of $$y$$ in $$[c,d]$$

How to proceed with the proof of the above theorem?

If an f(x,y) is continuous in a closed and bounded region, then f(x,y) is also uniformly continous here, so that setting...

$\displaystyle G(y) = \int_{a}^{b} f(x,y)\ dx\ (1)$

... for any h>0 is...

$\displaystyle |G(y + h) - G(y)| = | \int_{a}^{b} \{ f(x,y+h) - f(x,y)\}\ dx| \le \int_{a}^{b} |f(x,y+h) - f(x,y)|\ d x\ (2)$

Now f(x,y) is uniformly continuous so that choosing h 'small enough' You can do the last term of (2) 'small as You like' and that means that G(y) is continous...

Kind regards

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I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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