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

TheDestroyer

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

Hurkyl

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

TheDestroyer

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

I don't understand your objection: after integrating the Taylor series, you get the Taylor series for the result of the integral.

dextercioby

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

THAT was initially
[tex] \int \frac{\sin x}{x} dx [/tex]

Daniel.

P.S.Yours can be integrated exactly without any problem...

Zurtex

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

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

dextercioby

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

Daniel.

TheDestroyer

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

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.

dextercioby

They have antiderivatives,Si(x) and erf(ix) are (up until mulitplicative and additive constants) their antiderivatives...

Daniel.

dextercioby

Yes,part integration is a succesfull method of antidifferentiation,BUT NOT IN THIS CASE...

Daniel.

hedlund

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

Sure, hedlund:
This has already been mentioned by Hurkyl, hinted at by Daniel, and Zurtex has provided a link to a wolfram page.

dextercioby

You woke up a bit too late.This series method had been discussed in the first posts of the thread

Daniel.

Manchot

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.