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integrating sinx /x

I have no idea how to go about this.. integrating by parts won't work and I can't think of a substitution.
Original post by StarvingAutist
I have no idea how to go about this.. integrating by parts won't work and I can't think of a substitution.


Limits?
Reply 2
Avatar for Zacken
Zacken
22
Original post by StarvingAutist
I have no idea how to go about this.. integrating by parts won't work and I can't think of a substitution.


You'll need limits, it doesn't have an antiderivative in terms of elementary functions (other than its trivial Si(x)).
Original post by ODES_PDES
Limits?


70.png
Reply 4
Avatar for Alexion
Alexion
17
Original post by StarvingAutist
I have no idea how to go about this.. integrating by parts won't work and I can't think of a substitution.


It's a result in itself.

https://en.wikipedia.org/wiki/Trigonometric_integral
Original post by StarvingAutist
70.png


reverse the order of integration
Original post by ODES_PDES
reverse the order of integration


What should that look like?
Reply 7
Avatar for TeeEm
TeeEm
19
Original post by StarvingAutist
What should that look like?

...
IMG_20160517_0001.jpg274.8KB
Original post by TeeEm
...


I didn't realise the sinx / x could go in the y integral; I put it as a limit which obviously didn't help
Original post by StarvingAutist
70.png


Start by doing the y integral and it becomes very simple. On another note I did just work out the exact solution to the integral of sinx/x - if someone fancies checking, that would be great.

∫sin(x)xdx=−1∑i=0∞n!x−(n+1)eix+(−1)ne−ix22incos(π2(2n+12−1))+constant\displaystyle\int\frac{sin(x)}{x}dx = -1 \sum_{i=0}^\infty \frac{n!}{x^{-(n+1)}}\frac{{e^{ix}} + (-1)^ne^{-ix}}{2\sqrt{2}i^ncos(\frac{\pi}{2}(\frac{2n+1}{2}-1))} + constant
(edited 7 years ago)
Reply 10
Avatar for Zacken
Zacken
22
Original post by natninja
Start by doing the y integral and it becomes very simple. On another note I did just work out the exact solution to the integral of sinx/x - if someone fancies checking, that would be great.

∫sin(x)xdx=−1∑i=0∞n!x−(n+1)eix+(−1)ne−ix22incos(π2(2n+12−1))+constant\displaystyle\int\frac{sin(x)}{x}dx = -1 \sum_{i=0}^\infty \frac{n!}{x^{-(n+1)}}\frac{{e^{ix}} + (-1)^ne^{-ix}}{2\sqrt{2}i^ncos(\frac{\pi}{2}(\frac{2n+1}{2}-1))} + constant


Looks like you've done far more work that needs to be.

sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!\displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}

Then divide that by xx and integrate term-by-term (skipping over some mild technical justifications).

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