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Integrals can sometimes be computed through their saddle points. Are there examples of converse, when saddle points are found via integrals?
OK, but I do not assume that the moments are given. Rather, I'm interested to look at the situation when the measure is known explicitly, it's critical points are hard to find, but the moments can still be computed.
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Integrals can sometimes be computed through their saddle points. Are there examples of converse, when saddle points are found via integrals?
I think the difference with my question is that I assume that the measure $\mu$ is known, but it's difficult to find its critical points directly.
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Integrals can sometimes be computed through their saddle points. Are there examples of converse, when saddle points are found via integrals?
Well for real exponent the integral clearly localizes near the minimum. For complex exponent, as in you original computation, both critical points $t=\pm\frac12$ probably contribute and the analysis is more involved.
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Integrals can sometimes be computed through their saddle points. Are there examples of converse, when saddle points are found via integrals?
Replacing $i x$ by $x$ seems to fix the issue.
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Integrals can sometimes be computed through their saddle points. Are there examples of converse, when saddle points are found via integrals?
@CarloBeenakker I'm not sure why you say the integrals do not converge. Mathematica's
With[{x = 10}, NIntegrate[Exp[ x Sin[\[Pi] t]] t, {t, -1, 1}]/ NIntegrate[Exp[ x Sin[\[Pi] t]], {t, -1, 1}]]
gives me 0.499989
.
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Integrals can sometimes be computed through their saddle points. Are there examples of converse, when saddle points are found via integrals?
I didn't check carefully but it must be that (note additional $t$ in the numerator) $\int_{-1}^1 dt e^{i x \sin \pi t} t/\int_{-1}^1 dt e^{i x \sin \pi t}$ tends to $\frac12$ in the large $x$ limit. In this case of course one can find the saddle point directly. The procedure should always work (formally), my question is if there are any examples when this is useful.
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Theta-function in the lower half-plane
Hi, thanks for an elaborate answer! Unfortunately I can not recall my motivation vividly enough, but I think I was playing around with expressing integrals as sums over residues. I noticed that sometimes the contour can not be closed (because the function does not decay where it would need to) and the sum of residues diverges, but regularized sum gives the right answer for original finite integral. So the sum in OP is can probably interpreted as a formal sum of residues for some converging integral.
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Nonlinear ODE to linear PDE?
Thanks for both links, very instructive. However I had something a bit different in mind, perhaps not articulated clearly. As far as I can tell the paper that you linked mostly considers 1-1 mappings of between linear and non-linear diff equations (with some reservations for not 1-1 mappings.) The mappings exists only for very particular types of equations. In contrast I was asking if one can trade non-linearity for linearity in a possibly much larger space. In my example $q(t)$ is a one-parameter function with a non-linear equation while $\psi(q,t)$ is now two-parameter but obeying the linear
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Naive generalization of determinant from matrices to higher rank tensors
@MoritzEissler Thanks for the link! The question there is indeed the same, but the answer just states the formula from my post without giving it a name of explaining why is that a good definition. Another suggestion was to check out hyperdeterminant, but as far as I can see it is different from my formula.
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Decomposition of a sum of holomorphic squares into modular forms
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Decomposition of a sum of holomorphic squares into modular forms
Well, this question is CFT-inspired, but not exactly. I think you can show in general that if $f_i$ are linearly independent, they must furnish some representation of $PSL(2,\mathbb{Z})$. In CFT these things come as characters. Now I ask a reverse question: having such a decomposition, how can the space of functions $f_i$ be described? I don't even know the answer when $N=1$, so perhaps I should've started with a simpler question, I will make the corresponding edits.
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