# Another integral involving a Gaussian and a logarithm

By generalizing methods used in An integral involving a Gaussian and a logarithm. I have computed the following integral below: \begin{eqnarray} \tilde{\mathcal I}(A) &:=& \int\limits_{-1/A}^\infty \log(1+A \xi) \frac{e^{-\xi^2/2}}{\sqrt{2\pi}} d\xi \\ &=& \frac{1}{2} \log(\sqrt{2} A)+\frac{1}{4} \psi(1/2) + \frac{1}{4 A^2} F_{2,2}\left[\begin{array}{rr} 1 & 1 \\ 3/2 & 2\end{array};-\frac{1}{2 A^2}\right] + \frac{\log(\sqrt{2} A)-\gamma/2}{A \sqrt{2 \pi}} F_{1,1}\left[\begin{array}{r} 1/2 \\ 3/2 \end{array};-\frac{1}{2 A^2}\right] -1/2 \frac{1}{A \sqrt{2\pi}} F_{1,1}^{(1,0,0)}\left[\begin{array}{r} 1/2 \\ 3/2\end{array};-\frac{1}{2 A^2}\right] \end{eqnarray} Here $\psi$ denotes the di-gamma function and $F_{1,1}^{(1,0,0)}$ is the derivative of the hypergeometric function with respect to its first parameter. Clearly $\tilde{\mathcal I}(A) \simeq -A^2/2$ as $A \rightarrow 0_+$ and $\tilde{\mathcal I}(A) \simeq 1/2 \log(A)$ as $A \rightarrow \infty$.

The quantity $\tilde{\mathcal I}(A)$ along with the real part of the quantity ${\mathcal I}(A)$ are plotted below in Red and in Blue respectively.

Now, again the question is how will the result look like is we replace the Gaussian by a Tsallis' or a L\'{e}vy stable density function ?

• Isn't this a duplicate? – Did Jul 17 '15 at 9:46
• Both integrals are different and are computed using different ways. You can clearly see the the new integral isn't just the real part of the old one. The new integral has a derivative of the hypergeometric function whereas the old one didn't. Try to copy the derivation of the old integral and apply it to the new one -- I bet you won't be able to do it the old way. – Przemo Jul 17 '15 at 9:52
• @Did: Besides, I would really love to know how to compute the integral in question when the Gaussian is replaced by a L`{e}vy stable distribution. Note that in that case the second moment does not exist so the integral is not differentiable with respect to $A$ at $A=0$. – Przemo Jul 17 '15 at 10:08
• Isn't this integral the real part of the other one? – Did Jul 17 '15 at 10:09
• And the other question says: "Now the question is how would you calculate the integral in question if the Gaussian was replaced by a Tsallis' distribution", so what is new here: Tsallis+Lévy instead of Tsallis? – Did Jul 17 '15 at 10:10

Here we provide an answer for the Tsallis' case only. Denoting the unknown integral by $\tilde{\mathcal I}_q$ we clearly have(see the answer to the other question for justification): $$\tilde{\mathcal I}_q(A) = \frac{1}{\Gamma(\frac{1}{q-1}-\frac{1}{2})} \int\limits_0^\infty \frac{d s}{s} s^{\frac{1}{q-1}-\frac{1}{2}} e^{-s} \tilde{\mathcal I}\left(\frac{A}{\sqrt{s (q-1)}}\right)$$ Therefore inserting the right hand side of the equation in the formulation of the question into the above equation and integrating term by term just the same way as we did in the answer to the old question we get: \begin{eqnarray} \tilde{\mathcal I}_q(A) = \frac{1}{2} \log(\sqrt{2} A)-\frac{1}{4}\left(\log(q-1) + \psi(\frac{1}{q-1}-\frac{1}{2})\right) +\frac{1}{4} \psi(\frac{1}{2})+ \frac{1}{4 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}\\\frac{3}{2}&2\end{array};-\frac{(q-1)}{2 A^2}\right]+ \frac{{\mathcal A}_q}{A \sqrt{2 \pi}} (\log(A) -\frac{1}{2}(\log(q-1) + \psi(\frac{1}{q-1}))+(\log(\sqrt{2}) -\gamma/2))F_{2,1}\left[\begin{array}{rr}\frac{1}{2}& \frac{1}{q-1} \\ \frac{3}{2} \end{array};-\frac{(q-1)}{2 A^2}\right] - \frac{{\mathcal A}_q}{2 A \sqrt{2 \pi}} \left( \left. \partial_a F_{2,1}\left[\begin{array}{rr}a & \frac{1}{q-1} \\ \frac{3}{2} \end{array};-\frac{(q-1)}{2 A^2}\right] \right|_{a=1/2} + \left. \partial_a F_{2,1}\left[\begin{array}{rr}a & \frac{1}{2} \\ \frac{3}{2} \end{array};-\frac{(q-1)}{2 A^2}\right] \right|_{a=\frac{1}{q-1}} \right) \end{eqnarray} where $${\mathcal A}_q := \left( \frac{\sqrt{q-1}\Gamma(\frac{1}{q-1})}{\Gamma(\frac{1}{q-1}-\frac{1}{2})}\right)$$ As we can see the result has a nice symmetry; see the last two terms for example. This symmetry was not visible in the Gaussian case. It is usually the case that there is a need to generalize results because the generalized results are more symmetric than the simple ones. In other words the beauty reveals itself after an appropriate generalization only. Since ${\mathcal A}_q \rightarrow 1$ when $q\rightarrow 1_+$ and $(\log(q-1) + \psi(\frac{1}{q-1})) \rightarrow 0$ when $q\rightarrow 1_+$ we clearly see that $\tilde{\mathcal I}_q \rightarrow \tilde{\mathcal I}$ when $q$ goes to unity.
Here we provide the answer for a symmetric (ie unskewed) L\'{e}vy stable distribution with mean zero. Denote the PDF and the CDF of the stable distribution by $\rho_\mu(\xi)$ and by $\Phi_\mu(\xi)$ respectively.Then we have: \begin{eqnarray} &&\tilde{\mathcal I}_\mu(A) := \int\limits_{-1/A}^\infty \log(1+A \xi) \rho_\mu(\xi) d\xi \nonumber\\ &&= \frac{1}{2} \log(A) + \gamma \frac{1-\mu}{2 \mu} + \left(\log(A) - \gamma\right) \left( \Phi_\mu(\frac{1}{A})-\frac{1}{2} \right) + \nonumber\\ && \frac{1}{\pi \mu} \left[ -\frac{\pi}{2} \sum\limits_{n=1}^\infty \frac{(1/A)^n}{n!} \cos\left(\frac{\pi}{2}n\right) \Gamma\left(\frac{n}{\mu}\right) -\frac{1}{\mu} \sum\limits_{n=1}^\infty \frac{(1/A)^n}{n!} \sin\left(\frac{\pi}{2}n\right) \Gamma\left(\frac{n}{\mu}\right) \Psi\left(\frac{n}{\mu}\right) \right] \end{eqnarray} where $\Psi$ is the di-gamma function. The derivation follows according to the same lines as in the Gaussian case except that since we do not have an closed form expression for the PDF we express the PDF as an inverse Fourier transform of a stretched exponential and repeat the calculations from the Gaussian case.