I recently stumbled upon an indefinite integral . sin(x)/x [ Another similar one is root (x) times sin x. However if we substitute sin(x) in terms of x as Maclaurin series we could get a series of infinite yet integrable polynomials . What's the catch?
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1Then you're left with a sum of infinite terms that has no compact closed form. So you define one. – user170231 Jan 11 '19 at 15:50
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It's not solvable in terms of elementary functions. – snowfall512 Jan 11 '19 at 15:51
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Okay I get that fact . – Mikhail Tal Jan 11 '19 at 15:59
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Ok I get it ....but the thing is it's mentioned as "unintegrable" . – Mikhail Tal Jan 11 '19 at 15:59
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And what about the root(x) sin(x) one? – Mikhail Tal Jan 11 '19 at 15:59
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Who mentions that and what exactly do they say? – Robert Israel Jan 11 '19 at 16:16
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$$\int \sqrt{x}\sin(x); dx = - \sqrt{x} \cos(x) + \sqrt{\frac{\pi}{2}} \text{FresnelC}\left(\sqrt{\frac{2x}{\pi}}\right)$$ Again, this can't be expressed in elementary functions. – Robert Israel Jan 11 '19 at 16:20
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@Robert Israel It's mentioned in Class 12 NCERT Mathematics Books Chapter 7 Integrals page 324 bottom. – Mikhail Tal Jan 11 '19 at 16:22
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The thing is I guess it can be expressed in elementary polynomial functions...just an infinite series – Mikhail Tal Jan 11 '19 at 16:22
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A series doesn't count as an "elementary function". – Robert Israel Jan 11 '19 at 16:23
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God how could I miss that . – Mikhail Tal Jan 11 '19 at 16:25
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Robert can you please give your view on NCERT book the page I mentioned as well? Maybe they referred unintegrable in a different sense ? – Mikhail Tal Jan 11 '19 at 16:26
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Page 38 of this PDF : https://www.google.com/url?sa=t&source=web&rct=j&url=http://ncert.nic.in/ncerts/l/lemh201.pdf&ved=2ahUKEwizlujtkubfAhUIWX0KHZ8oA_wQFjAEegQIAxAB&usg=AOvVaw0x8ON34xX-Tuvq9r7RS3KM – Mikhail Tal Jan 11 '19 at 16:29
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Possible duplicate: 3065762/155 – user170231 Jan 11 '19 at 18:09
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The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $\int \sqrt{x} \sin\; x \; dx$. The reason is that there does not exist any function whose derivative is $\sqrt{x} \sin \; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $\sqrt{x} \sin\; x$. Integration by parts "works" on $\sqrt{x} \sin\; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ \int \sqrt{x} \sin \; x \; dx = \frac{2}{3} x^{3/2} \sin\; x - \frac{2}{3} \int x^{3/2} \cos\; x\; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
Robert Israel
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