How to integrate Sin(x)/(x)?


by TheDestroyer
Tags: integrate, sinx or x
TheDestroyer
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#1
Feb24-05, 03:19 PM
P: 390
Hi guys, I think the question is clear (lol)

How can the Sin(x)/(x) function be integrated? i heared it can be integrated using a series, anyone can explain?

Last added : I remembered the function e^(x^2) how also can it be integrated?

Thanks,

TheDestroyer
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Hurkyl
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#2
Feb24-05, 03:25 PM
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Can you think of a series for (sin x)/x?
TheDestroyer
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#3
Feb24-05, 03:29 PM
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Actually I studied taylor series, and using it will not give the general wanted answer as a function, but I heared it can be solved using the fourier series, I don't know, I really completely don't know what to do about it!!!!

Hurkyl
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#4
Feb24-05, 03:46 PM
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How to integrate Sin(x)/(x)?


I don't understand your objection: after integrating the Taylor series, you get the Taylor series for the result of the integral.
dextercioby
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#5
Feb24-05, 03:48 PM
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According to my ancient Maple,it is a constant times [itex] Si(x) [/itex] + a constant.For some authors (like the ones who produced Maple),the first constant is +1...There are other conventions,though,tipically "normalization" ones,v.Erf(x) ...

Daniel.
dextercioby
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#6
Feb24-05, 05:00 PM
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THAT was initially
[tex] \int \frac{\sin x}{x} dx [/tex]

Daniel.

P.S.Yours can be integrated exactly without any problem...
Zurtex
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#7
Feb24-05, 05:45 PM
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There are functions created (and used in some circles of mathematics) which basically mean the integrals you are asking:

Sine Integral: http://mathworld.wolfram.com/SineIntegral.html

Imaginary error function: http://mathworld.wolfram.com/Erfi.html
mathwonk
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#8
Feb24-05, 10:07 PM
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when you say "integrated" do you mean "antidifferentiated"?
dextercioby
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#9
Feb25-05, 03:51 AM
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Of course,what else,he wants to find the antiderivative for those 2 functions...

Daniel.
TheDestroyer
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#10
Feb25-05, 09:01 AM
P: 390
then there is not antiderivative for them !! even with a series?
dagger32
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#11
Feb25-05, 09:54 AM
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try integration by parts...that is usefull for evaluating an integral composed of two fuctions, in this case, your first integral include the fuctions [tex] \sin {x} [/tex] and [tex]\frac {\11}{x}[/tex] and your second integral includes [tex] e^x [/tex] and [tex] x^2[/tex]

Remember, integration by parts formula yeilds:


[tex] {u}{v} - \int{v}{du}[/tex]

and for the [tex]{e^{x^2}}[/tex] fuction, when you integrate by parts, make sure you set u equal to the fuction whose derivative will eventually go to zero, otherwise you will have a mess on your hands.
dextercioby
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#12
Feb25-05, 10:18 AM
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They have antiderivatives,Si(x) and erf(ix) are (up until mulitplicative and additive constants) their antiderivatives...

Daniel.
dextercioby
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#13
Feb25-05, 10:20 AM
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Quote Quote by dagger32
try integration by parts...that is usefull for evaluating an integral composed of two fuctions, in this case, your first integral include the fuctions [tex] \sin {x} [/tex] and [tex]\frac {\11}{x}[/tex] and your second integral includes [tex] e^x [/tex] and [tex] x^2[/tex]

Remember, integration by parts formula yeilds:


[tex] {u}{v} - \int{v}{du}[/tex]

and for the [tex]{e^{x^2}}[/tex] fuction, when you integrate by parts, make sure you set u equal to the fuction whose derivative will eventually go to zero, otherwise you will have a mess on your hands.
Yes,part integration is a succesfull method of antidifferentiation,BUT NOT IN THIS CASE...

Daniel.
hedlund
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#14
Feb25-05, 12:34 PM
P: 34
Can't you integrate sin(x)/x by using the fact that sin(x) = x - x^3/3! + x^5/5! - x^7/7! ... so sin(x)/x = 1 - x^2/3! + x^4/5! - x^6/7! ... this would give x - x^3/(3*3!) + x^5/(5*5!) - x^7/(7*7!) + C, and we know that sin(x)/x -> 1 as x -> 0, so C=1. This would give us that the antiderivate of sin(x)/x is:

1 + x - x^3/(3*3!) + x^5/(5*5!) - x^7/(7*7!) ...

And this can be expressed as an infinite sum if you like
arildno
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#15
Feb25-05, 12:51 PM
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Sure, hedlund:
This has already been mentioned by Hurkyl, hinted at by Daniel, and Zurtex has provided a link to a wolfram page.
dextercioby
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#16
Feb25-05, 12:51 PM
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You woke up a bit too late.This series method had been discussed in the first posts of the thread

Daniel.
Manchot
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#17
Feb26-05, 02:39 PM
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If you'd like a "better" series than a Taylor series, you might want to know the asymptotic expansion for the function. (By "better", I mean more, faster convergence).

[tex]\int{\frac{\sin{x}}{x}} = -\frac{\cos{x}}{x}-\int{\frac{\cos{x}}{x^2}}
= -\frac{\cos{x}}{x}-\frac{\sin{x}}{x^2}-2\int{\frac{\sin{x}}{x^3}}[/tex]

Notice that with each successive integration by parts, the remainder term gets smaller for large values of x. Thus, for all x above a certain value, this series should converge.
jeebus_on_steroids
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#18
Apr28-05, 11:13 AM
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you need to use the midpoint rule, the trapezium rule and then apply these to simpsons rule silly, what's the width ur integrating?


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