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I'm looking for a solution to the following integral. However, it seems it doesn't have a solution.

$$\int\limits_{\theta-1}^{x} \left(\frac{y+1-\theta}{\theta} \right)^{-\frac{\displaystyle1}{\displaystyle \epsilon+3}} y^{-\frac{\displaystyle\epsilon+4}{\displaystyle\epsilon+3}}dy,$$

where $\theta \ge 1$, $\epsilon \ge 0$, $x > \theta -1$ and it depends on a number of other parameters.

This equation appears in the context of Physical Layer Security, which is an area of study in digital communications (telecom).

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  • $\begingroup$ What is the range of $x, \theta, \epsilon$? It would also be helpful to know the context in which this integral came up, and if estimates are useful if a closed-form solution isn't possible. $\endgroup$
    – Dispersion
    Dec 28, 2021 at 18:05
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    $\begingroup$ Is $x>\theta -1$ ? A closed form solution means not very much in general: it is much more useful to know how to compute it numerically for practical purposes, or study its dependance on various elements. So what do you really want to know : how to evaluate it efficiently (for which range of parameters) or to know its behavior for $\theta$ close to $x+1$, or near $1$, or for $x$ large...? $\endgroup$
    – username
    Dec 28, 2021 at 18:51
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    $\begingroup$ the "solution" can be written in terms of a hypergeometric function, in view of the indefinite integral $$\int y^a(y+b)^c\,dy=(a+1)^{-1}y^{a+1} (b+y)^c \left(y/b+1\right)^{-c} \, _2F_1\left(a+1,-c;a+2;-y/b\right)$$ no idea if this useful for you... $\endgroup$ Dec 28, 2021 at 18:58
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    $\begingroup$ Letting $\alpha=\theta-1$, if $|x-\alpha|<\alpha$ (i.e. if $x/\alpha$ is within the unit-ball centered at $x=1$), then you have a binomial series expansion: $$\alpha^{-1-f(\epsilon)}\sum_{k\ge 0}(-1)^k \alpha^{-k} {k+f(\epsilon)\choose k} \frac{(x-\alpha)^{k-f(\epsilon)+1}}{k-f(\epsilon)+1},$$ where $f(\epsilon)=\frac{1}{3+\epsilon}$. $\endgroup$
    – Dispersion
    Dec 28, 2021 at 19:15
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    $\begingroup$ The expansion does work, although perhaps not if $x/\alpha$ is too large (remember that $\alpha=\theta-1$). As I stated, the expansion is valid for $|x-\alpha|<\alpha$, i.e for $x\in(0, 2\alpha)$, so even if $x>\alpha$, the series converges as long as $x\in(\alpha, 2\alpha)$. This result is of course derived within a certain limit, so it won't be valid for all $x$ such that $x\ge \alpha$; asking to know the exact behavior of your integral not in any limit but for any values of your parameters is not very reasonable given your integral. $\endgroup$
    – Dispersion
    Dec 29, 2021 at 19:24

1 Answer 1

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With some effort (the lower integration limit requires care) I found this answer for the definite integral:

$$I(x)=\int\limits_{\delta}^{x} \left(y-\delta\right)^{-\frac{1}{ \epsilon+3}} y^{-\frac{\epsilon+4}{\epsilon+3}}dy=\frac{ (-2/\delta)^{\frac{2}{{\varepsilon}+3}} \pi ^{3/2} }{\sin \left(\pi\frac{{\varepsilon}+2}{{\varepsilon}+3}\right)\Gamma \left(\frac{{\varepsilon}+1}{2 {\varepsilon}+6}\right) \Gamma \left(\frac{\varepsilon+4}{{\varepsilon}+3}\right)}$$ $$\qquad\qquad+\;\delta^{-1}({\varepsilon}+3) x^{-\frac{1}{{\varepsilon}+3}} (x-{\delta})^{\frac{{\varepsilon}+2}{{\varepsilon}+3}} \, _2F_1\left(1,\frac{{\varepsilon}+1}{{\varepsilon}+3};\frac{{\varepsilon}+2}{{\varepsilon}+3};\frac{x}{{\delta}}\right),$$ for $\delta\equiv \theta-1>0$ and $x>\delta$, $\varepsilon>0$. Each of the two terms on the right-hand-side is complex, but the imaginary parts cancel.


Here is a Mathematica command to test this result against a numerical evaluation of the integral:

Plot[{((-4)^(1/(3 + eps)) delta^(-(2/(3 + eps))) [Pi]^(3/2) Csc[((2 + eps) [Pi])/(3 + eps)])/( Gamma[(1 + eps)/(6 + 2 eps)] Gamma[ 1 + 1/(3 + eps)]) + ((3 + eps) x^(-(1/(3 + eps))) (-delta + x)^(( 2 + eps)/(3 + eps)) Hypergeometric2F1[1, (1 + eps)/(3 + eps), (2 + eps)/(3 + eps), x/ delta])/delta,
NIntegrate[ y^(-(eps + 4)/(eps + 3))*(y - delta)^(-1/(eps + 3)), {y, delta, x}]}, {x, delta, 2}]

with this output for $\varepsilon=0.2,\delta=0.3$ (two indistinguishable curves)

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