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### 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: \int_\mathbb{R} \...
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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 $$f(x) ... 0answers 106 views ### Show improperly Riemman integrable function is not Lebesgue integrable [duplicate] Let$$f(x)=\frac{\sin\left(\frac{1}{t}\right)}{t}.$$Show that \int^1_0 f(x)\,dx exists but f\notin L^1, i.e., on (0,1) f is improperly Riemann integrable but not Lebesgue integrable. ... 0answers 59 views ### Absolute integrability [duplicate] Bartle has a statement: Although the absolute value of a proper Riemann integrable function is Riemann integrable, this may no longer be the case for a function which has an improper Riemann ... 2answers 38 views ### Test the convergence of an improper integral [duplicate] Test the convergence of \int_0^∞\frac {\sin x}{x}\,dx. My attempt = by comparison test the integrand diverges but how it is conditionally convergent I don't understand. I think it diverges in both ... 0answers 36 views ### Prove or disprove the convergence of improper integral [duplicate] Prove or disprove that$$\int_0^\infty \left|\frac{\sin{x}}{x}\right|dx < \infty.$$I tried by Wolframalpha and this integral seems to satisfy the Cauchy property, but I do not know how to prove ... 28answers 95k views ### Evaluating the integral \int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2? A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:$$\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$Well, can ... 8answers 4k views ### Why do we restrict the definition of Lebesgue Integrability? The function f(x) = \sin(x)/x is Riemann Integrable from 0 to \infty, but it is not Lebesgue Integrable on that same interval. (Note, it is not absolutely Riemann Integrable.) Why is it we ... 3answers 5k views ### How to prove absolute summability of sinc function? We know that$$\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$$... 2answers 6k views ### Does Riemann integrable imply Lebesgue integrable? Suppose a definite integral exists in the Riemann sense. Does that mean the integral exists as a Lebesgue integral, and do we get the same result either way? ------- BTW: I have a MS in Electrical ... 3answers 5k views ### What exactly is a Non Integrable function? How can they be solved? I was trying to find the integral of \frac{\sin x}{x} recently, and nothing I tried appeared to get me any closer to a solution, so I looked it up and apparently it's a Non-integrable function. ... 3answers 2k views ### Problems with \int_{1}^{\infty}\frac{\sin x}{x }dx convergence I'd love your help with deciding whether the following integral converges or not and in what conditions: \int_{1}^{\infty}\frac{\sin x}{x}. 1. First, I wanted to use Dirichlet criterion: let f,g: [... 2answers 2k views ### How to prove that \frac{\sin x}{x} is not Lebesgue integrable on [0,+\infty]? How to prove that \displaystyle\int_0^{+\infty}\left|\dfrac{\sin x}{x}\right| \, dx = +\infty ? Could any one give some hint ? Thanks. 1answer 998 views ### Show that \int\nolimits^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2 [duplicate] Show that \int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2 by integrating z^{-1}e^{iz} around a closed contour \Gamma consisting of two portions of the real axis, from -R to -\epsilon and from ... 1answer 351 views ### How does one show that \int_0^\infty \left|\frac{\sin x} x\right| \, dx=\infty? [duplicate] In many place one finds accounts of how to evaluate$$ \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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