Linked Questions

2
votes
3answers
989 views

Test for convergence [duplicate]

Possible Duplicate: Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral? I am stuck with the following integral: \begin{equation} \int_\mathbb{R} \...
2
votes
1answer
895 views

Proving $f(x) = \frac{\sin x}{x}$ converges by improper integral test? [duplicate]

So my professor talked about one example of improper integrals and I'm having difficulty understanding the general proof outline for proving convergence. I was given the problem to prove that $$f(x) ...
0
votes
0answers
106 views

Show improperly Riemman integrable function is not Lebesgue integrable [duplicate]

Let $$f(x)=\frac{\sin\left(\frac{1}{t}\right)}{t}.$$ Show that $\int^1_0 f(x)\,dx$ exists but $f\notin L^1$, i.e., on $(0,1)$ $f$ is improperly Riemann integrable but not Lebesgue integrable. ...
1
vote
0answers
59 views

Absolute integrability [duplicate]

Bartle has a statement: Although the absolute value of a proper Riemann integrable function is Riemann integrable, this may no longer be the case for a function which has an improper Riemann ...
0
votes
2answers
38 views

Test the convergence of an improper integral [duplicate]

Test the convergence of $\int_0^∞\frac {\sin x}{x}\,dx$. My attempt = by comparison test the integrand diverges but how it is conditionally convergent I don't understand. I think it diverges in both ...
0
votes
0answers
36 views

Prove or disprove the convergence of improper integral [duplicate]

Prove or disprove that $$\int_0^\infty \left|\frac{\sin{x}}{x}\right|dx < \infty.$$ I tried by Wolframalpha and this integral seems to satisfy the Cauchy property, but I do not know how to prove ...
207
votes
28answers
95k views

Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?

A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$ Well, can ...
43
votes
8answers
4k views

Why do we restrict the definition of Lebesgue Integrability?

The function $f(x) = \sin(x)/x$ is Riemann Integrable from $0$ to $\infty$, but it is not Lebesgue Integrable on that same interval. (Note, it is not absolutely Riemann Integrable.) Why is it we ...
9
votes
3answers
5k views

How to prove absolute summability of sinc function?

We know that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi}{2}.$$ How do I show that $$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert dx$$ ...
10
votes
2answers
6k views

Does Riemann integrable imply Lebesgue integrable?

Suppose a definite integral exists in the Riemann sense. Does that mean the integral exists as a Lebesgue integral, and do we get the same result either way? ------- BTW: I have a MS in Electrical ...
2
votes
3answers
5k views

What exactly is a Non Integrable function? How can they be solved?

I was trying to find the integral of $\frac{\sin x}{x}$ recently, and nothing I tried appeared to get me any closer to a solution, so I looked it up and apparently it's a Non-integrable function. ...
5
votes
3answers
2k views

Problems with $\int_{1}^{\infty}\frac{\sin x}{x }dx$ convergence

I'd love your help with deciding whether the following integral converges or not and in what conditions: $\int_{1}^{\infty}\frac{\sin x}{x}$. 1. First, I wanted to use Dirichlet criterion: let $f,g: [...
5
votes
2answers
2k views

How to prove that $\frac{\sin x}{x}$ is not Lebesgue integrable on $[0,+\infty]$?

How to prove that $\displaystyle\int_0^{+\infty}\left|\dfrac{\sin x}{x}\right| \, dx = +\infty$ ? Could any one give some hint ? Thanks.
4
votes
1answer
998 views

Show that $\int\nolimits^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$ [duplicate]

Show that $\int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$ by integrating $z^{-1}e^{iz}$ around a closed contour $\Gamma$ consisting of two portions of the real axis, from -$R$ to -$\epsilon$ and from ...
4
votes
1answer
351 views

How does one show that $\int_0^\infty \left|\frac{\sin x} x\right| \, dx=\infty$? [duplicate]

In many place one finds accounts of how to evaluate $$ \int_0^\infty \frac{\sin x} x\,dx = \underbrace{\lim_{a\to\infty}\int_0^a}_{\text{Why view it this way?}} \frac{\sin x} x\, dx. $$ And it gets ...

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