How would you do it? Part 2
Let S = r = 1 ∑ ∞ ∫ − ∞ ∞ x 2 − 2 ( 2 ( r + 1 ) ) x + 3 r d x
Then find, S 3 + 9 S + 2 7
Note: Work with the Cauchy principal value of the integrals.
If you're looking for another challenge in calculus, try this
The answer is 27.
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2 solutions
I do not think we need contour integration here, i have simply used completing square and then integrated it to get of the form
ln(x-a/x+a) which clearly tends to ln(1) = 0 for both + and - infinity,
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Yup, it does. I did the same. @Mvs Saketh for the win.
Actually, the problem with that is that this function is discontinuous (at the roots, the denominator becomes zero).
Therefore, you have to integrate this piece-wise and can't directly make that claim. Also, try asking wolfram alpha to integrate this kind of integral and it will tell you that it diverges. But it will tell you that the 'Cauchy principle value' is zero. This happens because the contour chosen is only a limiting case.
@Mvs Saketh @Raghav Vaidyanathan
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oh yes, while going from - to + infinity, it is undefined at x = a and infinity at x=-a (again undefined), hence i cannot make that claim, but can i not claim that if very close to x= -a , the integral treated as an infinite sum has a value of ∞ d x and then at x=a it will be − ∞ d x which cancel out to be 0, as they are both the same type infinities (limits of the same varriable) , if this sounds a bit irrational, i can replace the infinities wth some varriable 'b' and tend it to infinity and then -b and tend it to -infinity
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@Mvs Saketh – You are subtracting two infinities and your claim that they are equal is based on the premise that you take two e q u i d i s t a n t limiting points on either side of the point of discontinuity.
This is precisely the idea behind the Cauchy principle value(which can be extended to multiple dimensions based on essentially the same idea). But obviously, this is an ad-hoc assumption and therefore may or may not be true in general.
This question and more importantly the discussion in the answers should help clarify matters further.
@Mvs Saketh @Raghav Vaidyanathan
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@Shashwat Shukla – Yes, i understand, it is true for a limiting case but that doesnt imply it will be true at that exact point as well unless we define it in a universally self consistent way . Or basically, this is maths and not physics, we need to be more rigorious , and i have seen the problem, that helped clear some idea, thanks :)
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@Mvs Saketh – Precisely :)...It's just a matter of definition.
It is not true that ∫ − ∞ ∞ x − d 1 d x = 0 because the integral diverges at x = d . See https://brilliant.org/problems/inspired-by-pranjal-jain/ for example.
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@Jon Haussmann ...Thank you so much for bringing my mistake to my attention... It was a blatant error on my part.
I actually saw a proof of the result(for the integral of the quadratic being zero) that uses contour integration. I thought that this makes for an easier proof...well, it doesn't...I have made appropriate changes.
Only r = {1, 2, 3, 4, 5} are hopeful with possible value of integrals as others tend to zeroes before infinities are substituted in. Therefore, we try to consider only first 5 to sum up for S.
x^2 - 16 x + 3
x^2 - 256 x + 9
x^2 - 65536 x + 27
x^2 - 4294967296 x + 81
x^2 - 18446744073709551616 x + 343
All these are having same form of integral:
0.5/ Sqrt (b) [Ln (1 - 2 Sqrt (b)/ (x - a + Sqrt (b)))] for x from negative infinity to positive infinity = 0.5/ Sqrt (b) [Ln 1 - Ln 1] = 0. We find that this is also true for r = {6, 7, 8 , 9, ...} Despite seeming greater, it is checked and determined that the integrals are getting smaller actually. Therefore,
S = 0
S^3 + 9 S + 27 = 27
There is a more expedient approach to arrive at the same conclusion. Do check out my solution.
The point of this question is to bring to your attention the rather neat looking property: ∫ − ∞ ∞ a x 2 + b x + c 1 d x = 0 ∀ a , b , c ∈ ℜ w h e r e b 2 > 4 a c i.e. If the quadratic in the denominator has real roots. The fact that the discriminant of all the quadratics in this question are positive, is easily verified.
All the complicated expressions introduced in the problem were just noise: an attempt to discourage "Wolfram-alpha-ing" or guesswork.
The property in question isn't as easy to prove as may seem at first sight: It can be proven by contour integration, but I omit the proof here.
By use of the result, all the terms of the summation in the given question are zero and the final answer is just 2 7 .