
#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

Sci Advisor
HW Helper
P: 9,398

is that an indefinite or definite integral?




#3
Feb1804, 08:24 PM

Math
Emeritus
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Thanks
PF Gold
P: 38,881

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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P: 9,398

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




#7
Feb1904, 10:02 AM

Sci Advisor
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P: 2,538

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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P: 2,538

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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P: 2,538

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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P: 9,398

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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P: 9,398

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

Sci Advisor
HW Helper
P: 2,538

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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