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something that I found confusing me. Lets see for example the follow integral:

$$\int_{A}^{B}e^{t\cdot x} \cdot dx=\left[\frac{e^{t\cdot x}}{t}\right]_{A}^B=\frac{e^{t\cdot B}-e^{t\cdot A}}{t}$$

From this results we may conclude that for $t=0$ , the integral is undefined: $$\frac{e^{t\cdot B}-e^{t\cdot A}}{t}=\frac{1-1}{0}=\frac{0}{0}$$

but from the other hand, if we evaluate the integral again for this case of $t=0$, then we get a defined results: $$\int_{A}^{B}e^{t\cdot x} \cdot dx=\int_{A}^{B}e^{0\cdot x} \cdot dx=\int_{A}^{B}1\cdot dx=B-A$$

So is this integral is defined for $t=0$ or not?

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2 Answers 2

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In the $\frac{0}{0}$ case, we can apply L'Hôpital's rule to obtain $$lim_{t\to 0}\frac{e^{tB}-e^{tA}}{t}=lim_{t\to 0}\frac{Be^{tB}-Ae^{tA}}{1}=B-A$$ so the two results agree. If we integral a continuous function of two variables with respect to one variable, the result should be a continuous function of the other.

In general, if $F(t)=\int_a^b f(t,x)dx$, we would define, for fixed $t_0$, $F(t_0)=\int_a^b f(t_0,x)dx$ rather than evaluating for general $t$ and then setting $t=t_0$. But assuming $f$ is sufficiently nice, them the two will agree so long as we take limits

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  • $\begingroup$ Thanks! regarding the $\frac{0}{0}$ case - the problem for me is for when $t=0$, not when it tends $lim_{t\to 0}$. Am I wrong with that ? $\endgroup$
    – audi02
    Mar 30, 2020 at 19:54
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You are using the formula $\int e^{tx}\, dx = \frac{e^{tx}}{t} + C$, but this rule is not valid when $t=0$. You should actually use

$$\int e^{tx}\, dx = \begin{cases} \frac{e^{tx}}{t} + C &\text{ if } t \neq 0 \\ x + C &\text{ if } t = 0.\\ \end{cases}$$

Then the two results are the same. In general, when you use a formula involving a division be aware that the formula will not be valid when the denominator is $0$. This is often an unstated assumption, so you should check if both sides a given equation are undefined or not in that case.

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  • $\begingroup$ Great. now we see also for $t=0$, and that they are fit: when $lim_{t\to 0}$ , it tends to B-A, and when $t=0$ it is B-A. This example that two different parts are fit, raises an interesting question: Can we know initially if all function integral parts should fit? Or it is here just "by accident". $\endgroup$
    – audi02
    Mar 31, 2020 at 5:58
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    $\begingroup$ An integral identity should be valid for everywhere both sides of the equation are defined. If the formula for the evaluation of the integral is not defined at a given point, you'll have to think through whether the integral can still be evaluated at that point. Technically, this case should be included in the formula, but sometimes it may be left off depending on the source. $\endgroup$ Mar 31, 2020 at 17:03

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