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 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.

[dextercioby]

THAT was initially
\int \frac{\sin x}{x} dx :wink:

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 \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.

[dextercioby]

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

Daniel.

[dextercioby]

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... :wink:

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 :tongue:

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).

\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.

[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?

[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

[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.

[Mr.Rabbit87]

well.... i looked through all the replies to this question and felt that no one really answered it. sooooo

i got the integral to sin(x)/(x) to = -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c

done by integration by parts twice.

so integral of udv = uv - integral of vdu

u = lnx
du = 1/x
dv = sinx
v = -cosx

u get -cosx/x + (integral of cosxlnx) ------ do integration by parts again

u = lnx
du = 1/x
dv = cosx
v = sinx

so u get lnxsinx - (integral of sinx/x). add the (integral of sinx/x) over. So now you have 2(integral of sinx/x) = -cosx/x + lnxsinx

divide the 2 over n you get -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c

[Mr.Rabbit87]

well.... i looked through all the replies to this question and felt that no one really answered it. sooooo

i got the integral to sin(x)/(x) to = -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c

done by integration by parts twice.

so integral of udv = uv - integral of vdu

u = lnx
du = 1/x
dv = sinx
v = -cosx

u get -cosx/x + (integral of cosxlnx) ------ do integration by parts again

u = lnx
du = 1/x
dv = cosx
v = sinx

so u get lnxsinx - (integral of sinx/x). add the (integral of sinx/x) over. So now you have 2(integral of sinx/x) = -cosx/x + lnxsinx

divide the 2 over n you get -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c

i lied... i differentiate wrong lol