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

by TheDestroyer
Tags: integrate, sinx or x
 P: 397 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
 Emeritus Sci Advisor PF Gold P: 16,091 Can you think of a series for (sin x)/x?
 P: 397 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!!!!
 Emeritus Sci Advisor PF Gold P: 16,091 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.
 Sci Advisor HW Helper P: 11,928 According to my ancient Maple,it is a constant times $Si(x)$ + 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.
 Sci Advisor HW Helper P: 11,928 THAT was initially $$\int \frac{\sin x}{x} dx$$ Daniel. P.S.Yours can be integrated exactly without any problem...
 Sci Advisor HW Helper P: 1,123 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
 Sci Advisor HW Helper P: 9,488 when you say "integrated" do you mean "antidifferentiated"?
 Sci Advisor HW Helper P: 11,928 Of course,what else,he wants to find the antiderivative for those 2 functions... Daniel.
 P: 397 then there is not antiderivative for them !! even with a series?
 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 $$\sin {x}$$ and $$\frac {\11}{x}$$ and your second integral includes $$e^x$$ and $$x^2$$ Remember, integration by parts formula yeilds: $${u}{v} - \int{v}{du}$$ and for the $${e^{x^2}}$$ 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.
 Sci Advisor HW Helper P: 11,928 They have antiderivatives,Si(x) and erf(ix) are (up until mulitplicative and additive constants) their antiderivatives... Daniel.
HW Helper
P: 11,928
 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 $$\sin {x}$$ and $$\frac {\11}{x}$$ and your second integral includes $$e^x$$ and $$x^2$$ Remember, integration by parts formula yeilds: $${u}{v} - \int{v}{du}$$ and for the $${e^{x^2}}$$ 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.
 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
 Sci Advisor HW Helper PF Gold P: 12,016 Sure, hedlund: This has already been mentioned by Hurkyl, hinted at by Daniel, and Zurtex has provided a link to a wolfram page.
 Sci Advisor HW Helper P: 11,928 You woke up a bit too late.This series method had been discussed in the first posts of the thread Daniel.
 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). $$\int{\frac{\sin{x}}{x}} = -\frac{\cos{x}}{x}-\int{\frac{\cos{x}}{x^2}} = -\frac{\cos{x}}{x}-\frac{\sin{x}}{x^2}-2\int{\frac{\sin{x}}{x^3}}$$ 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.
 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?

 Related Discussions Calculus 22 General Math 6 Calculus 8 Calculus & Beyond Homework 4 Calculus & Beyond Homework 5