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

1. Feb 24, 2005

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

Last edited: Feb 24, 2005
2. Feb 24, 2005

Hurkyl

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

3. Feb 24, 2005

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!!!!

4. Feb 24, 2005

Hurkyl

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

5. Feb 24, 2005

dextercioby

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.

6. Feb 24, 2005

dextercioby

THAT was initially
$$\int \frac{\sin x}{x} dx$$

Daniel.

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

7. Feb 24, 2005

Zurtex

8. Feb 24, 2005

mathwonk

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

9. Feb 25, 2005

dextercioby

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

Daniel.

10. Feb 25, 2005

TheDestroyer

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

11. Feb 25, 2005

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.

Last edited: Feb 25, 2005
12. Feb 25, 2005

dextercioby

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

Daniel.

13. Feb 25, 2005

dextercioby

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

Daniel.

14. Feb 25, 2005

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

15. Feb 25, 2005

arildno

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

16. Feb 25, 2005

dextercioby

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

Daniel.

17. Feb 26, 2005

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).

$$\int{\frac{\sin{x}}{x}} = -\frac{\cos{x}}{x}-\int{\frac{\cos{x}}{x^2}} = -\frac{\cos{x}}{x}-\frac{\sin{x}}{x^2}-2\int{\frac{\sin{x}}{x^3}}$$

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.

18. Apr 28, 2005

jeebus_on_steroids

you need to use the midpoint rule, the trapezium rule and then apply these to simpsons rule silly, what's the width ur integrating?

19. Mar 9, 2011

kasunjbandara

open your MATLAB and insert the operation
then differentiate the answer given by MATLAB
then search the solution reverse....
(start from the final step and come to the first step)

sometimes the above method will work...but even I am not sure about it

20. Mar 9, 2011

djsourabh

There could be other way as to go for fourier transform keeping f=0 (frequency )in its equation.If h(x)=\hat{f}(x) then  \hat{h}(\xi)= f(-\xi). that is duality of fourier transform.