Convergence and Integration: The Role of Constant C and the Limit of Functions

fk378
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When representing a function as a power series, why do we evaluate the constant C of integration at 0 to determine the value of C?

Also, if the limit of a function is 0, does that mean that the function itself converges to 0?
 
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fk378 said:
When representing a function as a power series, why do we evaluate the constant C of integration at 0 to determine the value of C?
What does "representing a function as a power series" have to do with integrating? The most common way of writing a function as a power series is the Taylor's series method that involves differentiation. Could you give a specific example of what you are talking about?

Also, if the limit of a function is 0, does that mean that the function itself converges to 0?
What do you mean by "function itself converges to 0"? "Convergence" always implies a limit. Do you mean the value of the function is 0?

If the limit, as x goes to a, of f(x) is any specific value L (which could be 0) and f is continuous at a, then, yes, f(a)= L. That is the definition of "continuous".
 
For the first question I am just referring to when figuring out the constant C from an indefinite integral, is the protocol to plug in zero into f(x) to see what the value of C is?
 
Not always, 0 may be the most convenient but any point within the domain of the function can be used. Sometimes 0 is not in the domain.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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