Value of the integral $\int_{a}^{a}f(x)dx?$ I am confused about the value of the integral $\int_{a}^{a}f(x)dx,$ is it $0$ or not always? In Riemann integration case (i.e. $f$ is bounded ) its zero by definition. What about other cases for example $\int_{0}^{0}\frac{1}{x}? $ Is it zero? Please help. Thanks a lot.
 A: It's always zero...except that there are some conventions in which this notation is used to denote something that's not zero. 
So let me clarify "always": for the Riemann integral, and for the  Lesbegue integral with respect to the standard measure on the real line, and for functions $f: \mathbb R \to \mathbb R$, it's always zero. This property is very useful, for it helps ensure that 
$$
\int_a^b + \int_b^c = \int_a^c
$$
in general. 
So why do I say "except..."?
Well, for instance, if $f$ is a "delta function" (which isn't really a function, but instead a distribution), then this integral can denote something nonzero (or at least
$$
\int_{0^{-}}^{0^{+}} f(x) dx
$$
can be nonzero). 
And if your measure "dx" on the real line happens to assign positive measure to the point $0$, then again the integral can denote something nonzero. But it'd be pretty odd to use "dx" to denote such a measure. 
A: The heart of your question comes down to what we mean by the word integral. The most reasonable value for $\int_{0}^{0}\frac{1}{x}\,\mathrm{d}x$ is probably $0$, and for this reason nobody would object to you giving the value of this integral as $0$. I expound on this below.
Let $a<b$ be real numbers and let $f\colon[a,b]\to\mathbb{R}$. Broadly speaking, if $\int_{c-\epsilon}^{c+\epsilon}f(x)\,\mathrm{d}x$ is defined for all sufficiently small $\epsilon>0$, then we usually would define $\int_{c}^{c}f(x)\,\mathrm{d}x$ to be zero. This is simply because defining it any other way would be strange: our intuitive picture of what the integral means is "an area under a graph". It then makes intuitive sense that, if the area under the graph is a meaningful concept for some small interval around $c$ (that is, if the function is integrable on $[c-\epsilon,c+\epsilon]$ for some $c\in\mathbb{R}$ and some $\epsilon>0$), then the area under the graph (the value of the integral) should get smaller and smaller (tend to zero) as we make the interval smaller (as $\epsilon\to0$).
Ultimately, a fair amount of what is true in pure mathematics comes down to what we define to be true. We make the definitions ourselves according to what we consider to be "reasonable". Whether or not the statement above is actually true for any particular definition of "integral" depends on the definitions, either of the integral itself or just of integrals with these kinds of limits. My point is that most "reasonable" definitions of "integral" are called "reasonable" at least in part because this statement is true for them. If it weren't, they wouldn't (usually) be very good definitions of "integral".
This solves the problem in cases where the function is sufficiently integrable. In other cases, such as the example $\int_{0}^{0}\frac{1}{x}\,\mathrm{d}x$ that you give, whether or not the value is $0$ is again dependent on what we define it to be. If we wanted to, we could define $\int_{0}^{0}\frac{1}{x}\,\mathrm{d}x=\pi$ and see where that gets us, but we would never want to do this because this makes little sense. The most reasonable value would be $0$, for a couple of reasons.


*

*The integrand is an odd function, and we expect from our knowledge of "nice" situations that the integral of an odd function on an interval symmetric about $0$ should evaluate to $0$. Since $\{0\}$ is in a perverse sense "an interval symmetric about $0$", it makes sense to define the value of the integral based on this other result.

*A related way to come up with a reasonable definition is to take inspiration from other reasonable definitions, in this case the Cauchy principal value. The Cauchy principal value allows us to assign values to certain improper integrals by imagining the area under the graph as "expanding out from the origin". In this case, we may instead imagine the area as expanding in from outside the origin. With this picture in mind, we might define
$$\int_{0}^{0}\frac{1}{x}\,\mathrm{d}x = \lim_{\epsilon\to1^{-}}\left(\int_{-1}^{-1+\epsilon}\frac{1}{x}\,\mathrm{d}x + \int_{1-\epsilon}^{1}\frac{1}{x}\,\mathrm{d}x=0\right).$$
Now the calculation is easy: for any $\epsilon\in(0,1)$ we have $\int_{-1}^{-1+\epsilon}\frac{1}{x}\,\mathrm{d}x + \int_{1-\epsilon}^{1}\frac{1}{x}\,\mathrm{d}x=0$ by symmetry (either evaluate the integral or draw a diagram). Hence the limit is $0$.


The important thing to take away here is that if something is undefined, you may sometimes feel free to define it yourself; the important thing is that the definition should be mathematically reasonable. Of course, what is reasonable and what isn't reasonable is another question entirely! Figuring out the right answer to this question is often the main goal of much mathematical research.
A: You can say like this $$\int_0^0{1\over x}dx=\lim_{c\to0+}\int_{-c}^{c}{1\over x}dx=\lim_{c\to0+}\left[\ln|x|\right]_{-c}^c=\lim_{c\to 0+}\{\ln|c|-\ln|-c|\}=0$$
