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I am unable to understand the following question.
Is the integral $$\int^\frac{π}{2}_0 \frac{\sin x}{x} \,\mathrm{d}x$$ improper and why?
Also what is an impromper integral?

Did
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Farouq
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  • No noninformative titles, please. – Did Mar 15 '18 at 09:30
  • This will be a question in a particular textbook, regarding the exact definitions in that textbook. I can easily imagine different textbooks with different technical definitions of "improper integral" regarding this point. – GEdgar Mar 15 '18 at 12:59
  • @Faroq Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 17 '18 at 23:18

2 Answers2

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Since $\frac{sin(x)}{x} \to 1$ as $x \to 0$, we can define $f: \mathbb R \to \mathbb R$ by

$f(x):=\frac{sin(x)}{x}$ if $x \ne 0$ and $f(0):=1$. Then $f$ is continuous and $ \int_0^{\pi/2}\frac{sin(x)}{x} dx = \int_0^{\pi/2}f(x) dx$.

Consequence: $ \int_0^{\pi/2}\frac{sin(x)}{x} dx $ is not improper.

Fred
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Without any other information the integral $\int^\frac{π}{2}_0 \frac{\sin x}{x} \,\mathrm{d}x$ is to be considered improper since $\frac{\sin x}{x}$ is not defined for $x=0$ and we need to define it by limit

$$\int^\frac{π}{2}_0 \frac{\sin x}{x} \,\mathrm{d}x=\lim_{a\to 0^+} \int^\frac{π}{2}_a \frac{\sin x}{x} \,\mathrm{d}x$$

and it converges since $x\to 0$ $\frac{\sin x}{x}\to1$.

But it is not improper if we define the integrand $\frac{\sin x}{x}=a\in \mathbb{R}$ for $x=0$ and in particular $a=1$ which leads to a continuos integrand function; in this case indeed we are not forced to define the integral by limit.

user
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  • Sorry but I’m confused by your answer. – Farouq Mar 15 '18 at 10:26
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    @Faroq By definition an improper integral is the limit of a definite integral as an endpoint (or both in some case) of the interval of integration approaches either a specified real number or $\infty$ https://en.wikipedia.org/wiki/Improper_integral . In the first case since $\frac{\sin x}{x}$ is not defined we need to consider the integral by limit definition thus it is an improper integral but if we eliminate this undefined point we can consider the definite integral. – user Mar 15 '18 at 10:36
  • @Faroq I've assumed "a" since one single point of discontinuity doesn't affect the integral but of course we can assume a=1 if we like to have a continuos integrand function. – user Mar 15 '18 at 10:36
  • Thank you for your answer – Farouq Mar 15 '18 at 11:23
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    @Faroq You are welcome, I hope it and the Others answers given have helped you to solve your doubts, in this scase remember that you can upvote and evaluate if accept an answer among the given, more detials here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 15 '18 at 11:26
  • if you mind , can you help me with the other question that I posted in another post which is about finding unknown value of c ? – Farouq Mar 15 '18 at 12:25