Behavior of a function defined by two infinite integrals around the singularity point

Question

Bounty ending tomorrow:

Consider the following function defined by two infinite integrals: $$ F(\epsilon) = \int_0^\infty \frac{q^3 (1+q)e^{-2q}}{q^4+\epsilon^4} \, \mathrm{d} q + \int_0^\infty \frac{\, G_q \, e^{-q}}{q^2(q^4+\epsilon^4)} \, \mathrm{d} q $$ where $$ G_q := G \left( \left[ \left[\frac{1}{2} \right] , [\,]\right] , \left[ \left[ \frac{9}{2},\frac{5}{2} \right], \left[\frac{3}{2}\right] \right] ,\frac{q^2}{4} \right) \, , $$ with $G$ being the Meijer G-function. The goal is to study the behavior of $F$ around the singularity point, i.e. as $\epsilon\to 0$. Numerically, it can clearly be observed that $F$ scales logarithmically with $\epsilon$. I am wondering whether this behavior can be shown analytically via a rigorous analysis, e.g. using perturbation techniques. Your help or hints are highly appreciated and are most welcome.

Thanks H

Answer 1

Your Integral may be expressed as a Meijer G-Function of two Variables, which is discussed by Agarwal in 1964 for the first time. A more general form is the H-Fox-Function of two Variables which can be found in the standard publication of Mathai: https://www.researchgate.net/publication/266566090_The_H-function_Theory_and_Applications.

The first of the two indefinite integrals:

$$F_{1}\left( \epsilon \right) =\int_{0}^{\infty }\frac{q^{3}\left( 1+q\right) \exp \left( -2q\right) }{q^{4}+\epsilon ^{4}}dq$$

can be solved analytically in terms of two Meijer G-functions:

$$F_{1}\left( \sigma \right) \frac{1}{16\sqrt{2}\pi ^{\frac{3}{2}}\sigma ^{3}}% \left[ G_{1,5}^{5,1}\left( 16\,\sigma ^{4}\left\vert \begin{array}{c} -,-,\frac{3}{4},-,- \\ \frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{7}{4}% \end{array}% \right. \right) +\frac{1}{4\,\sigma }G_{1,5}^{5,1}\left( 16\,\sigma ^{4}\left\vert \begin{array}{c} -,-,1,-,- \\ 1,1,\frac{5}{4},\frac{3}{2},\frac{7}{4}% \end{array}% \right. \right) \right] $$

where we use $\sigma =\frac{\epsilon }{4}$ to have the same notation later on for the second indefinite integral:

$$F_{2}\left( \epsilon \right) =\int_{0}^{\infty }\frac{1}{q^{2}\left( q^{4}+\epsilon ^{4}\right) }G_{1,3}^{2,1}\left( \frac{q^{2}}{4}\left\vert \begin{array}{c} -,\frac{1}{2},- \\ \frac{9}{2},\frac{5}{2},\frac{3}{2}% \end{array}% \right. \right) \exp \left( -q\right) dq$$

The first and the last function can be found in The wolfram functions side. They can be expressed in terms of their Meijer G-identities and then $F_{2}$ can be evaluated using the integration theorem for Meijer G-functions (http://functions.wolfram.com/HypergeometricFunctions/MeijerG/21/02/04/)

$$\int_{0}^{\infty }t^{\alpha -1}G_{p,q}^{m,n}\left( \left\vert \begin{array}{c} a_{1},...,a_{p} \\ b_{1},...,b_{q}% \end{array}% \right. \right) \times $$

$$G_{p_{1},q_{1}}^{m_{1},n_{1}}\left( x\,t\left\vert \begin{array}{c} a_{11},...,a_{1p_{1}} \\ b_{11},...,b_{1q_{1}}% \end{array}% \right. \right) \times G_{p_{2},q_{2}}^{m_{2},n_{2}}\left( y\,t\left\vert \begin{array}{c} a_{21},...,a_{2p_{2}} \\ b_{21},...,b_{2q_{2}}% \end{array}% \right. \right) dt$$ $$z^{-\alpha }G_{p,q,p_{1},q_{1},p_{2},q_{2},}^{m,n,m_{1},n_{1},m_{2},n_{2}}\left( \begin{array}{c} 1-\alpha -b_{1},...,1-\alpha -b_{q} \\ 1-\alpha -a_{1},...,1-\alpha -a_{p}% \end{array}% \left\vert \begin{array}{c} a_{11},...,a_{1p_{1}} \\ b_{11},...,b_{1q_{1}}% \end{array}% \right. \left\vert \begin{array}{c} a_{21},...,a_{2p_{2}} \\ b_{21},...,b_{2q_{2}}% \end{array}% \right. \left\vert \frac{x}{z},\frac{y}{z}% \begin{array}{c} \\ \end{array}% \right. \,\right) $$

The Meijer G-identities corresponding to the functions are: $$\exp \left( -q\right) =G_{0,1}^{1,0}\left( q\left\vert \begin{array}{c} -,- \\ 0,-% \end{array}% \right. \right) $$ $$\frac{1}{q^{2}\left( q^{4}+\epsilon ^{4}\right) }=\frac{1}{\epsilon ^{6}}% G_{1,1}^{1,1}\left( \left( \frac{q}{\epsilon }\right) ^{4}\left\vert \begin{array}{c} -\frac{1}{2},- \\ -\frac{1}{2},-% \end{array}% \right. \right) $$

After some algebraic manipulations and variable transformations, using (http://functions.wolfram.com/HypergeometricFunctions/MeijerG/17/02/03/) we conclude:

$$F_{2}\left( \sigma \right) =\frac{1}{32\sqrt{2}\pi ^{\frac{5}{2}}\sigma ^{6}}\int_{0}^{\infty }t^{\frac{1% }{4}}G_{1,1}^{1,1}\left( \frac{t}{\sigma ^{4}}\left\vert \begin{array}{c} -\frac{1}{2},- \\ -\frac{1}{2},-% \end{array}% \right. \right) \times G_{2,6}^{4,2}\left( t\left\vert \begin{array}{c} -,-,\frac{1}{4},\frac{3}{4},-,- \\ \frac{9}{4},\frac{11}{4},\frac{5}{4},\frac{7}{4},\frac{3}{4},\frac{5}{4}% \end{array}% \right. \right) $$

$$\times G_{0,4}^{4,0}\left( t\left\vert \begin{array}{c} -,-,-,- \\ 0,\frac{1}{2},\frac{1}{4},\frac{3}{4}% \end{array}% \right. \right) dt =$$

$$= \frac{1}{32\sqrt{2}\pi ^{\frac{5}{2}}\sigma ^{5}}% ~G_{1,1,2,6,0,4}^{1,1,4,2,4,0}\left( \begin{array}{c} \frac{5}{4},- \\ \frac{5}{4},-% \end{array}% \left\vert \begin{array}{c} -,-,\frac{1}{4},\frac{3}{4},-,- \\ \frac{9}{4},\frac{11}{4},\frac{5}{4},\frac{7}{4},\frac{3}{4},\frac{5}{4}% \end{array}% \right. \left\vert \begin{array}{c} -,-,-,- \\ 0,\frac{1}{2},\frac{1}{4},\frac{3}{4}% \end{array}% \right. \left\vert \sigma ^{4},\sigma ^{4}% \begin{array}{c} \\ \end{array}% \right. \,\right) $$

A similar example can be found here: https://www.researchgate.net/publication/269504875_Capacity_of_k_-_m_Shadowed_Fading_Channels. You may find also the source code for implementing the function in Mathematica in the publication above. Due to the lack of time, I can not check, if the solution of your Integral reduces to functions of only one Variable, since $\frac{x}{z}=\frac{y}{z}=\sigma ^{4}$. You may do this numerical, also I can provide the H-Fox-Function of one variable.

Answer 2

A technique similar to this will work. The tail at infinity can be neglected, and we have $$ I_1 = \int_0^\infty \frac {q^3 (1 + q)} {q^4 + \epsilon^4} e^{-2q} dq \sim \int_0^1 \frac {q^3} {q^4 + \epsilon^4} dq \sim -\ln \epsilon.$$ The expansion for the G-function at zero is found from the residues: $$G_q = G_{1,3}^{2,1}\left( \frac {q^2} 4 \middle| {\frac 1 2 \atop \frac 5 2, \frac 9 2, \frac 3 2} \right) \sim\\ -\operatorname{Res}_{y=5/2} \frac {\Gamma\left( \frac 1 2 + y \right) \Gamma\left( \frac 5 2 - y \right) \Gamma\left( \frac 9 2 - y \right)} {\Gamma \left( - \frac 1 2 + y \right)} \left( \frac {q^2} 4 \right)^y = \frac {q^5} {16}.$$ Then, in the same way as for $I_1$, $$I_2 = \int_0^\infty \frac {G_q} {q^2 \left( q^4 + \epsilon^4 \right)} e^{-q} dq \sim \int_0^1 \frac {q^3} {16(q^4 + \epsilon^4)} dq \sim -\frac {\ln \epsilon} {16},\\ F(\epsilon) = I_1 + I_2 \sim -\frac {17 \ln \epsilon} {16}.$$ To obtain the next term, a regularization can be applied to $I_1$ to get $$I_1 \sim -\ln \epsilon + \int_0^1 \left( \frac {q^3 (1 + q)} {q^4 + \epsilon^4} e^{-2q} - \frac 1 q \right) \bigg\rvert_{\epsilon=0} dq + \int_1^\infty \frac {q^3 (1 + q)} {q^4 + \epsilon^4} e^{-2q} \bigg\rvert_{\epsilon=0} dq,$$ and, repeating the same procedure for $I_2$, we find $$F(\epsilon) \sim -\frac {17 \ln \epsilon} {16} - \gamma - \ln 2 + \frac 1 2 +\\ \int_0^1 \left( q^{-6} e^{-q} G_q - \frac 1 {16 q} \right) dq + \int_1^\infty q^{-6} e^{-q} G_q dq.$$