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Consider a following integral: \begin{equation} {\mathcal I}(A) := \int\limits_{\mathbb R} \log\left(1+ A \xi\right) \cdot \frac{\exp(-\frac{\xi^2}{2})}{\sqrt{2 \pi}} d\xi \end{equation} By using the trick $\left. d \xi^a/d a \right|_{a=0} = \log(\xi)$ then by substituting for $\xi^2/2$ and then by expanding the resulting power-law term into a binomial expansion and integrating the result term by term using the definition of the Gamma function I have shown that: \begin{equation} {\mathcal I}(A) = \log(\frac{A}{\sqrt{2}})+ \imath \frac{\pi}{2} - \frac{\gamma}{2} + \frac{1}{2 A^2} F_{2,2}\left[\begin{array}{rr} 1 & 1 \\ 2 & 3/2\end{array};-\frac{1}{2 A^2}\right]- \imath \frac{\sqrt{\pi}}{\sqrt{2} A} F_{1,1}\left[\begin{array}{r} \frac{1}{2} \\ \frac{3}{2} \end{array};-\frac{1}{2 A^2}\right] \end{equation} The real part of the function in question is plotted below: enter image description here

It clearly behaves as a parabola ${\mathcal I}(A) \simeq -A^2/2$ for $A \rightarrow 0$ and as a logarithm ${\mathcal I}(A) \simeq \log(A)$ for $A\rightarrow \infty$.

Now the question is how would you calculate the integral in question if the Gaussian was replaced by a Tsallis' distribution $\rho_q(\xi) := 1/C_q \cdot e_q(-1/2 \xi^2)$ where $e_q(\xi) := [1+(1-q) \xi]^{1/(1-q)}$ for $q> 1$ ? Here $C_q:=\left(\sqrt{2\pi}/\sqrt{q-1}\right) \cdot \Gamma\left(\frac{3-q}{2(q-1)}\right)/\Gamma\left(\frac{1}{q-1}\right)$ is a normalization constant.

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  • $\begingroup$ Not an answer, just a comment: $\log(1+A\xi)$ can be written as $\log(A)+\log(1/A+\xi)$. Then if you translate $\xi$ by $-1/A$, you can see your integral as a convolution by a Gaussian (or Tsallis' filter), i.e. a smoothed (low-passed) version of the logarithm of $-1/A$, plus the term $\log(A)$. $\endgroup$
    – user65203
    Commented Jul 14, 2015 at 14:24
  • $\begingroup$ @Yves Daoust: Thank you for that comment. By the way, if as you rightfully say, the integral can be expressed as a convolution maybe one can use Fourier transforms and the convolution theorem to get a result? $\endgroup$
    – Przemo
    Commented Jul 15, 2015 at 10:20
  • $\begingroup$ Unfortunately, the $\log$ doesn't have a Fourier transform, as far as I know, even though due to the fast damping of the Gaussian, the integral does exist. $\endgroup$
    – user65203
    Commented Jul 15, 2015 at 10:26

2 Answers 2

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Let us denote the unknown integral with the Tsallis' distribution ${\mathcal I}_q(A)$. In what follows we assume that $3 > q> 1$. Note that the Tsallis' distribution has a very nice integral representation. In fact it can be seen as a Gaussian whose inverse variance is Gamma distributed. To be precise the following equality holds: \begin{equation} \rho_q(\xi) = \frac{1}{\Gamma(\frac{1}{q-1}) C_q} \int\limits_0^\infty \frac{ ds}{s} s^{\frac{1}{q-1}} e^{-s} e^{-s (q-1) \frac{\xi^2}{2}} \end{equation} Now, replacing the Gaussian in the formulation of the question by $\rho_q(\xi)$ and swapping the order of integration we immediately see that: \begin{equation} {\mathcal I}_q(A) = \frac{1}{\Gamma(\frac{1}{q-1}-\frac{1}{2})} \int\limits_0^\infty \frac{d s}{s} s^{\frac{1}{q-1}-\frac{1}{2}} e^{-s} {\mathcal I}\left(\frac{A}{\sqrt{s (q-1)}}\right) \end{equation} Now, the formula for ${\mathcal I}(A)$ consists of four types of terms, firstly a constant$(A)$, secondly a a $\log(A)$, thirdly a negative even power of $A$ and finally a negative odd power of $A$. Replacing ${\mathcal I}$ by those terms in the above equation, we can clearly see (since the integrals are trivial) that except for the logarithm , the terms obtain a multiplicative factor as a result of the above transformation. The case of the logarithm is slightly more complicated since it involves the di-gamma function $\psi$. Bringing everything together we readily have: \begin{equation} {\mathcal I}_q(A) = \log(\frac{A}{\sqrt{2(q-1)}}) - \frac{1}{2} \psi(\frac{1}{q-1}-\frac{1}{2}) + \imath \frac{\pi}{2} - \frac{\gamma}{2}+ \frac{1}{2 A^2}\left(1-\frac{1}{2}(q-1)\right) F_{3,2}\left[\begin{array}{rrr}1 & 1 & \frac{1}{q-1}+\frac{1}{2} \\ 2 & 3/2 \end{array};-\frac{(q-1)}{2 A^2}\right] -\imath \sqrt{\pi} \frac{\sqrt{q-1}}{\sqrt{2} A} \frac{\Gamma(\frac{1}{q-1})}{\Gamma(\frac{1}{q-1}-\frac{1}{2})} F_{2,1}\left[\begin{array}{rr}1/2 & \frac{1}{q-1}\\3/2\end{array};-\frac{(q-1)}{2 A^2}\right] \end{equation} The quantity above clearly tends to ${\mathcal I}$ as $q\rightarrow 1$, as it should be.

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In here we derive the result for ${\mathcal I}(A)$. Clearly we have: \begin{equation} {\mathcal I}(A) = \left. \frac{d}{d a} J(a,A) \right|_{a=0} \end{equation} where \begin{equation} J(a,A) := \int\limits_{\mathbb R} \left(1+A \xi\right)^a \frac{e^{-\frac{\xi^2}{2}}}{\sqrt{2 \pi}} d\xi \end{equation} Now we compute the quantity ${\mathcal J}$. We have: \begin{eqnarray} {\mathcal J}(a,A) &=& \int\limits_0^\infty \left[(1+A \sqrt{2 u})^a + (1-A \sqrt{2 u})^a\right] e^{-u} \frac{d u}{\sqrt{u}} \frac{1}{2 \sqrt{\pi}}\\ &=& \frac{(A \sqrt{2})^a}{2\sqrt{\pi}} \int\limits_0^\infty \left[(1+\frac{1}{A \sqrt{2 u}})^a + (-1)^a(1-\frac{1}{A \sqrt{2 u}})^a\right]u^{a/2-1/2} e^{-u} du\\ &=& \frac{A^a 2^{a/2}}{2 \sqrt{\pi}} \sum\limits_{n=0}^\infty \binom{a}{n} \frac{1+(-1)^{a+n}}{(A \sqrt{2})^n} \Gamma(\frac{a}{2}+\frac{1}{2}-\frac{n}{2})\\ &=&\frac{(A \sqrt{2})^a}{2\sqrt{\pi}} a! \sum\limits_{n=0}^\infty \frac{1+(-1)^{a+n}}{(A \sqrt{2})^n} \frac{\Gamma(\frac{a+1-n}{2})}{\Gamma(a+1-n)}\\ &=& \frac{A^a 2^{-a/2}}{2 \pi} a! \sum\limits_{n=0}^\infty \frac{1}{n!} (\frac{\sqrt{2}}{A})^n \left(1+(-1)^{a+n}\right) \sin(\pi(a/2+1-n/2)) \Gamma(-a/2+n/2)\\ &=& \frac{a!}{2} (\frac{A}{\sqrt{2}})^a \left[ \frac{(1+(-1)^a)}{\Gamma(1+a/2)} F_{1,1}\left[\begin{array}{r} -a/2\\ 1/2 \end{array};-\frac{1}{2 A^2}\right] + \frac{(1-(-1)^a)}{\Gamma((1+a)/2)} \frac{\sqrt{2}}{A} F_{1,1}\left[\begin{array}{r} (1-a)/2\\ 3/2 \end{array};-\frac{1}{2 A^2}\right] \right] \end{eqnarray} The first two steps in the derivation above are self-explanatory. In the third step I used the binomial expansion and integrated term by term using the definition of the Gamma function. In the fourth step I simplified the result and in the fifth step I used the duplication formula for the Gamma function. Finally in the last step I split the sum into a sum over even and odd integers respectively and then used the definition of hypergeometric functions. Now all we need to do is to differentiate the result at $a=0$ by applying the chain rule. Let us analyze the first term first: \begin{eqnarray} \left.\frac{d}{d a} \left[\frac{1}{2} (\frac{A}{\sqrt{2}})^a \cdot (1+(-1)^a) \cdot \frac{\Gamma(1+a)}{\Gamma(1+a/2)}\cdot F_{1,1}\left[\begin{array}{r} -a/2\\ 1/2 \end{array};-\frac{1}{2 A^2}\right]\right] \right|_{a=0} = \\ \frac{1}{2} \log(\frac{A}{\sqrt{2}}) \cdot 2 \cdot 1 \cdot 1 + \\ \frac{1}{2} \cdot (\imath \pi) \cdot 1 \cdot 1 +\\ \frac{1}{2} \cdot 2 \cdot (-\gamma) \cdot 1 +\\ \frac{1}{2} \cdot 2 \cdot 1 \cdot \frac{1}{2 A^2} F_{2,2}\left[\begin{array}{rr} 1& 1 \\ 2 & 3/2\end{array};-\frac{1}{2 A^2}\right] \end{eqnarray} The second term is easier to analyze since there will be only one non-zero term. We have: \begin{eqnarray} \left.\frac{d}{d a} \left[\frac{1}{2} (\frac{A}{\sqrt{2}})^{a-1} \cdot (1-(-1)^a) \cdot \frac{\Gamma(1+a)}{\Gamma((1+a)/2)}\cdot F_{1,1}\left[\begin{array}{r} (1-a)/2\\ 3/2 \end{array};-\frac{1}{2 A^2}\right]\right]\right|_{a=0}=\\ \frac{1}{2} (\frac{A}{\sqrt{2}})^{-1} \cdot (-\imath \pi) \cdot \frac{1}{\sqrt{\pi}} \cdot F_{1,1}\left[\begin{array}{r} 1/2\\ 3/2 \end{array};-\frac{1}{2 A^2}\right] \end{eqnarray} This finishes the proof.

Update: Below i just attach a piece of code in Mathematica that demonstrates that the result is indeed correct. See below:

A = RandomReal[{0, 10}, WorkingPrecision -> 20];
NIntegrate[
 Log[1 + A t] Exp[-t^2/2]/Sqrt[2 Pi], {t, -Infinity, Infinity}, 
 Exclusions -> {-1/A}, WorkingPrecision -> 20]
Log[A/Sqrt[2]] + I Pi/2 - EulerGamma/2 + 
 1/(2 A^2) HypergeometricPFQ[{1, 1}, {2, 3/2}, -1/(2 A^2)] - 
 I Sqrt[Pi]/(Sqrt[2] A) Hypergeometric1F1[1/2, 3/2, -1/(2 A^2)]

enter image description here

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