
#1
Feb2405, 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 



#2
Feb2405, 03:25 PM

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PF Gold
P: 16,101

Can you think of a series for (sin x)/x?




#3
Feb2405, 03:29 PM

P: 390

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!!!!




#4
Feb2405, 03:46 PM

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PF Gold
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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.




#5
Feb2405, 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. 



#6
Feb2405, 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... 



#7
Feb2405, 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 



#8
Feb2405, 10:07 PM

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P: 9,428

when you say "integrated" do you mean "antidifferentiated"?




#9
Feb2505, 03:51 AM

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Of course,what else,he wants to find the antiderivative for those 2 functions...
Daniel. 



#10
Feb2505, 09:01 AM

P: 390

then there is not antiderivative for them !! even with a series?




#11
Feb2505, 09:54 AM

P: 4

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. 



#12
Feb2505, 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. 



#13
Feb2505, 10:20 AM

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Daniel. 



#14
Feb2505, 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 



#15
Feb2505, 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. 



#16
Feb2505, 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. 



#17
Feb2605, 02:39 PM

P: 728

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. 



#18
Apr2805, 11:13 AM

P: 1

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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