# Continous value of integrals

Let $$f:(a,b] \rightarrow \mathbb{R}$$ if the riemann Integral$$F(c):= \int_c^b f(x) dx$$ for all $$c \in (a,b)$$ exists and the improper integral $$F(a)=\int_a^b f(x) dx$$ exists too. Does this mean that F is continuous on [a,b]?

i am completely sure, that this is true, but i need to be sure about this.

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sorry, i just meant is –  Lipschitz Apr 20 at 12:28

$F$ must be continuous at $a$ pretty much by definition. $\int_a^b f(x)\ dx$ is defined as $\lim\limits_{c\to a} \int_c^bf(x)\ dx$. In other words, $F(a) = \lim\limits_{c \to a} F(c)$, which means precisely that $F$ is continuous at $a$.

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this is not the excercise, the question is, whether the VALUE OF THE INTEGRAL is a continuous function. –  Lipschitz Apr 20 at 12:54
@Lipschitz: sorry, I misunderstood. I'll fix it. –  Javier Badia Apr 20 at 12:57
Consider a counterexample: $$f:(0,1]\longrightarrow \mathbb R$$ $$f(x)=\dfrac{1}{x}$$
Then $\int_c^1f(x)dx=\int_c^1\frac{1}{x}dx=\log 1-\log c= -\log c$. Therefore, the integral exists for all $c\in (0,1)$.
However, $$\int _0^1\frac{1}{x}dx=\lim\limits_{c\rightarrow 0^+}\int_{c}^1\frac{1}{x}dx=\lim\limits_{c\rightarrow 0^+}(-\log c)=+\infty$$ Hence, the integral does not converge. This means that $F(c)=\int_c^1f(x)dx$ is not defined in $c=0$, so it cannot be continuous there.
This is just the same answer as the other one, already erased. It only proves $$\int\limits_0^1\frac{dx}{x}$$ diverges and thus is not counterexample at all to what the OP asked. –  DonAntonio Apr 20 at 14:04