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How the heck do I integrate sin(1/x) ? 
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#1
Feb1804, 06:06 PM

P: 80

I tried parts by integration but Im caught in an endless loop of ever growing in complexity integrals! I must be missing something.



#2
Feb1804, 06:21 PM

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P: 9,396

is that an indefinite or definite integral?



#3
Feb1804, 08:24 PM

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PF Gold
P: 39,682

Do you have any reason to believe its antiderivative is an elementary function?



#4
Feb1804, 11:46 PM

P: 640

How the heck do I integrate sin(1/x) ?
easy man
here's the answer: cosintegral[1/x] + xSin[1/x] 


#5
Feb1904, 06:47 AM

P: 80

I got that, too. No way to further simplify?



#6
Feb1904, 07:21 AM

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apart from that the integral of 1/x is log(x) you mean?



#7
Feb1904, 10:02 AM

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If you're desperate, you could try working out a Taylor/Mclaurin series for it, and seeing if the integral of that is recognizable.



#8
Feb2004, 04:12 PM

P: 186

You can use a Maclaurin series to evaluate (or at least approximate) it...knowing that
[tex]sin(x)=\sum_{n=0}^{\infty}\frac{(1)^nx^{(2n+1)}}{(2n+1)!}[/tex] you can replace x with 1/x and integrate to get: [tex]\int sin(\frac{1}{x})=\sum_{n=0}^{\infty}\frac{(1)^{n1}}{2(2n+1)!x^{2n}}[/tex] 


#9
Feb2004, 07:10 PM

P: 640

Wrong. ∫1/x dx = ln x + C. ∫1/(x(ln 10)) dx = log x + C. 


#10
Feb2004, 07:13 PM

P: 678




#11
Feb2004, 08:32 PM

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but anything other than [tex]log_e[/tex] gets a base. 


#12
Feb2004, 11:56 PM

P: 678




#13
Feb2104, 12:22 AM

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It's typically for math/cs tpe situations and usually only applies to situations where hte base is not particuarly important.



#14
Feb2104, 03:12 AM

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yes, i did omit the modulus sign, however you should probably be told that log always means base e. This is completely standard in mathematics, and just one more thing they misteach at high school After all what other base would you possibly want? 


#15
Feb2404, 11:12 AM

P: 20

This might help too:
sin(1/x) is an odd function (meaning f(x) = f(x)). The definite integral of any odd function on the interval [a,a] is 0. 


#16
Feb2404, 12:04 PM

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one generally wouldn't integrate over a region where the function is not defined. (no choice at zero can make it continuous, interestingly enough, not that that's either here or there, and not that any choice would make the integral be anything but zero anyway, though 0 is the only choice that keeps it a genuine odd function.)



#17
Feb2404, 05:17 PM

P: 20

I should have been more careful when answering, but isn't the integral still well defined since {0} is a set of measure 0?



#18
Feb2404, 05:28 PM

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P: 2,537

Do you mean to use Lebesgue integration?
[tex]\lim_{x \rightarrow 0}[/tex] might also not exist and thus cause problems. 


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