Register to reply 
How to integrate Sin(x)/(x)? 
Share this thread: 
#1
Feb2405, 03:19 PM

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 


#2
Feb2405, 03:25 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

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



#3
Feb2405, 03:29 PM

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!!!!



#4
Feb2405, 03:46 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

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

Sci Advisor
HW Helper
P: 11,896

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

Sci Advisor
HW Helper
P: 11,896

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

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 


#8
Feb2405, 10:07 PM

Sci Advisor
HW Helper
P: 9,453

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



#9
Feb2505, 03:51 AM

Sci Advisor
HW Helper
P: 11,896

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


#10
Feb2505, 09:01 AM

P: 397

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

Sci Advisor
HW Helper
P: 11,896

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


#13
Feb2505, 10:20 AM

Sci Advisor
HW Helper
P: 11,896

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

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. 


#16
Feb2505, 12:51 PM

Sci Advisor
HW Helper
P: 11,896

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?



Register to reply 
Related Discussions  
What is the best way to integrate this?  Calculus  22  
How do you integrate this?  General Math  6  
Integrate ln(4+y^2)dy?  Calculus  8  
How do I integrate this?  Calculus & Beyond Homework  4  
How to integrate?  Calculus & Beyond Homework  5 