Linked Questions

7
votes
5answers
3k views

Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral? [duplicate]

Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral? I can't find any this kind of solution. Can anyone please help me? Thank you.
2
votes
3answers
1k views

Evaluating $\lim_{b\to\infty} \int_0^b \frac{\sin x}{x}\, dx= \frac{\pi}{2}$ [duplicate]

Possible Duplicate: Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? Using the identity $$\lim_{a\to\infty} \int_0^a e^{-xt}\, dt = \frac{1}{x}, x\gt 0,$$ can I ...
3
votes
4answers
346 views

Prove that $\int_0^\infty \frac{\sin nx}{x}dx=\frac{\pi}{2}$ [duplicate]

There was a question on multiple integrals which our professor gave us on our assignment. QUESTION: Changing order of integration, show that $$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy=\...
10
votes
2answers
2k views

A Complex approach to sine integral [duplicate]

this integral: $$\int_0^{+\infty}\frac{\sin x}{x}\text{d}x=\frac{\pi}{2}$$ is very famous and had been discussed in the past days in this forum. and I have learned some elegant way to computer it. ...
4
votes
1answer
1k 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 ...
0
votes
1answer
4k views

Dirichlet integral. [duplicate]

I want to prove $\displaystyle\int_0^{\infty} \frac{\sin x}x \,\mathrm{d}x = \frac \pi 2$, and $\displaystyle\int_0^{\infty} \frac{|\sin x|}x \,\mathrm{d}x \to \infty$. And I found in wikipedia, but ...
1
vote
1answer
2k views

how to integrate $ \int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx $? [duplicate]

Possible Duplicate: Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? How can I do this integration using only calculus? (not laplace transforms or complex ...
2
votes
3answers
194 views

Confirm that $\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2$ [duplicate]

Confirm that $\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2$. The guide book I am using gives the following help: Consider $\int_{\gamma}z^{-1}e^{iz}dz$, where for $0<s<r<\infty$ the contour of ...
3
votes
1answer
501 views

A 'complicated' integral: $ \int \limits_{-\infty}^{\infty}\frac{\sin(x)}{x}$ [duplicate]

I am calculating an integral $\displaystyle \int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x}$ and I dont seem to be getting an answer. When I integrate by parts twice, I get: $$\displaystyle \int \...
0
votes
0answers
894 views

Improper integration of sin(z)/z [duplicate]

I need to calculate this $\int_{-\infty}^\infty \frac{sin(z)}{z}dz$ using the residue of $\frac{sin(z)}{z}$. Then I write $\int_{-\infty}^\infty \frac{sin(z)}{z}dz$ = $\lim_{R\to\infty}Im(\int_{-R}^R ...
-1
votes
1answer
685 views

How to prove $\int^{\infty}_{0}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$ [duplicate]

Possible Duplicate: Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? How to prove $$\int^{\infty}_{0}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$$ Do you have a ...
1
vote
1answer
252 views

How to prove $\operatorname{si}(0) = -\pi/2$ without contour [duplicate]

Possible Duplicate: Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? How to prove $\operatorname{si}(0) = -\pi/2$ without contour integration ? Where $\...
1
vote
2answers
236 views

Evaluate $\int\limits_0^{\infty}\frac{\sin x}{x}\, dx$ [duplicate]

How can I evaluate the following improper integral: $$ \int\limits_0^{\infty}\frac{\sin x}{x}\, dx$$ I have tried to evaluate this by integration by parts but failed. Please help
0
votes
1answer
266 views

How can I evaluate $\int_0^\infty \frac{\sin x}{x} \,dx$? [may be duplicated] [duplicate]

How can I evaluate $\displaystyle\int_0^\infty \frac{\sin x}{x} \, dx$? (Let $\displaystyle \frac{\sin0}{0}=1$.) I proved that this integral exists by Cauchy's sequence. However I can't evaluate ...
3
votes
1answer
323 views

Proving $\int_\mathbb R\frac{\sin(x)}{x}dx = \pi$ using the residue theorem [duplicate]

I've been searching the web for a way to prove that $\int^{\infty}_{-\infty}{\sin(x)/x} = \pi$ with complex analysis, because I have a problem of consistency. I found two, carried in the following ...

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