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Let $A>0$ and $1\le \mu \le 2$. Consider a following definite integral: \begin{equation} {\mathcal I}(A,\mu) := Re\left[\int\limits_0^\infty e^{-(k A)^\mu}\frac{\left(\gamma+\Gamma(0,\imath k)+\log(\imath k)\right)}{\imath k}\cdot\left(-1+\mu (k A)^\mu \log(k A)\right) dk\right] \end{equation} By using similar methods as those in Limit behavior of a definite integral that depends on a parameter. , ie by expanding a part of the integrand in a Taylor series about $k=0$ and integrating term by term , we have shown that: \begin{equation} {\mathcal I}(A,\mu) = \frac{1}{\mu^2} \sum\limits_{n=1}^\infty \frac{(1/A)^n}{n!} \sin(\frac{\pi}{2} n) \Gamma\left(\frac{n}{\mu}\right) \cdot \Psi\left(\frac{n}{\mu}\right) \end{equation} Here $\Psi$ is the di-gamma function. Unfortunately the series above does not converge for small values of $A$. Again, we ask the question what is the asymptotic behaviour of the function in question when $A$ goes to zero.

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  • $\begingroup$ I have checked numerically that for small values of $A$ the integral in question behaves as: \begin{equation} {\mathcal I}(A,\mu) = \frac{\pi}{2}\left(\log(A) - \gamma\right) + O\left(A^\mu\right) \end{equation} $\endgroup$
    – Przemo
    Commented Aug 7, 2015 at 15:56

2 Answers 2

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This is going to be a ''hand-waving'' answer to this problem, ie lacking mathematical rigor. However, we will see that it the result will be ''almost'' correct. Again, if $A$ is small then the integral can be written as: \begin{eqnarray} {\mathcal I}(A,\mu) = Re\left[\int\limits_0^{\frac{1}{A}} 1 \frac{\left(\gamma+\Gamma(0,\imath k)+\log(\imath k)\right)}{\imath k}\cdot\left(-1+\mu (k A)^\mu \log(k A)\right) dk\right] + {\mathcal J}(A,\mu) \end{eqnarray} where $\lim_{A \rightarrow 0_+} {\mathcal J}(A,\mu) < \infty$. Therefore the problem boils down to finding an antiderivative of the integrand above. It is not difficult to find that antiderivative, it is just a tedious task. The only non-trivial terms in here are those that involve the upper incomplete Gamma function. However those terms can be delt with by by using the integral representation of the Gamma function, swapping the order of integration and integrating elementary functions.The result is the following: \begin{eqnarray} I^{(0)}_{A,\mu}(k):=\int \frac{\left(\gamma+\Gamma(0,\imath k)+\log(\imath k)\right)}{\imath k}\cdot\left(-1+\mu (k A)^\mu \log(k A)\right) dk =\\ \mu A^\mu \left[ \left. \frac{d}{d a} f(k,a) \right|_{a=\mu} + \log(A) f(k,\mu) \right] - g(k) \end{eqnarray} where \begin{eqnarray} f(k,\mu) &=& \frac{1}{\mu \imath}\left[k^\mu\left(\gamma +\imath \frac{\pi}{2} + \log(k) -\frac{1}{\mu} + \Gamma(0,\imath k)\right) - \imath^{-\mu} \Gamma\left(\mu,\imath k\right)\right] \\ g(k) &=& \frac{1}{\imath}\left[\left(\gamma+\imath \frac{\pi}{2}\right)\log(k) + \frac{1}{2} \log(k)^2 +\left(\log(k) + \imath \frac{\pi}{2}\right) \Gamma(0,\imath k) - \left.\frac{d}{d a} \Gamma(a,\imath k)\right|_{a=0}\right] \end{eqnarray} We have verified with Mathematica that the derivative of $I^{(0)}_{A,\mu}(k)$ with respect to $k$ gives indeed the integrand in question. Now, by expanding the incomplete Gamma function in a Maclaurin series and then taking carefuly the limit $k\rightarrow 0$ we get: \begin{equation} I^{(0)}_{A,\mu}(0) = \frac{1}{24} \left(\frac{12 (A/i)^{\mu } \Gamma (\mu ) (2 i \mu \log (A)+\pi \mu +2 i (\mu \psi ^{(0)}(\mu )-1))}{\mu }+i( \pi ^2 -12 \gamma ^2) +12 \gamma \pi\right) \end{equation} We also have: \begin{eqnarray} &&I^{(0)}_{A,\mu}(1/A) = \\ &&\frac{1}{2} i \left(2 i^{-\mu } A^{\mu } G_{2,3}^{3,0}\left(\frac{i}{A}| \begin{array}{c} 1,1 \\ 0,0,\mu \\ \end{array} \right)-2 G_{2,3}^{3,0}\left(\frac{i}{A}| \begin{array}{c} 1,1 \\ 0,0,0 \\ \end{array} \right)+\right. \\ &&\left. \frac{-2 e^{-\frac{1}{2} i \pi \mu } \mu A^{\mu } \Gamma \left(\mu ,\frac{i}{A}\right)+ 2 \mu \Gamma \left(0,\frac{i}{A}\right)+ } {\mu ^2}+\right.\\ &&\left. \frac{\log(A) ( \mu ^2 \log(A) - ( 2 \gamma \mu ^2 +i \pi \mu ^2 +2 \mu ) )+ 2\gamma \mu + i \pi \mu - 4}{\mu^2} \right) \end{eqnarray} where $G_{2,3}^{3,0}$ are Meier G functions (they are related to first derivatives of the incomplete Gamma function with respect to the first parameter). Now the integral reads: \begin{eqnarray} {\mathcal I}(A,\mu) &=& Re\left[I^{(0)}_{A,\mu}(1/A) - I^{(0)}_{A,\mu}(0) \right] + {\mathcal J}(A,\mu) \\ &=& \left(\frac{\pi}{2} \log(A) - \frac{\pi}{2 \mu}\right) - \frac{\pi \gamma}{2} + O\left(A^\mu\right) + {\mathcal J}(A,\mu) \end{eqnarray} Therefore the singular part of the integral reads: \begin{equation} \frac{\pi}{2}\left(\log(A)-\frac{1}{\mu}-\gamma\right) \end{equation} This is very close to the conjectured form (see comment to question). The result is only off by $-1/\mu$.

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Let us answer a simpler question first. We want to find a small-$A$ expansion of the following integral: \begin{equation} {\mathcal I}_1(A,\mu) := \int\limits_0^\infty e^{-(k A)^\mu} \cdot\left(\frac{\gamma+\Gamma(0, \imath k) + \log(\imath k)}{\imath k}\right) dk \end{equation} We will be constructing an antiderivative of the integrand and then taking it in the limits from zero to infinity. Below I quote two results: \begin{eqnarray} \int \left(\frac{\gamma+\Gamma(0, \imath k) + \log(\imath k)}{\imath k}\right) dk = \\\frac{1}{\imath}\left[(\gamma+\imath \frac{\pi}{2}) \log(k) + \frac{1}{2} \log(k)^2 - \left.\frac{d}{d a} \Gamma(a,\imath k) \right|_{a=0} + (\log(k) + \imath \frac{\pi}{2}) \Gamma(0,\imath k)\right] \\ \int k^{n \mu} \left(\frac{\gamma+\Gamma(0, \imath k) + \log(\imath k)}{\imath k}\right) dk =\\ \frac{1}{\imath}\left[(\gamma+\imath \frac{\pi}{2}) \frac{k^{n\mu}}{n \mu} + \frac{k^{n \mu}(-1+n\mu \log(k))}{(n \mu)^2} - \frac{\imath^{-n\mu}}{n\mu} \Gamma(n\mu,\imath k) + \frac{k^{n\mu}}{n\mu} \Gamma(0,\imath k)\right] \end{eqnarray} The results are not difficult to obtain. The nontrivial integral is the one that involves the upper incomplete Gamma function and some power function. That integral is easily done by using the integral representation of the Gamma function, swapping the order of integration and using integrals of elementary functions.

Now we quote the anti-derivative of the whole integrand.We have: \begin{eqnarray} \int e^{-(k A)^\mu} \cdot \left(\frac{\gamma+\Gamma(0, \imath k) + \log(\imath k)}{\imath k}\right) dk = \\ \frac{1}{\imath} \frac{(A k)^\mu}{\mu^2} F_{3,3} \left[\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}; -(A k)^\mu\right] - \frac{1}{\imath} \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{n\mu}}{n \mu} \Gamma(n\mu,\imath k) \\ +\frac{1}{\imath} \left(\frac{\gamma + \Gamma(0,(A k)^\mu) + \log((A k)^\mu)}{\mu}\right)\left(-\gamma - \imath \frac{\pi}{2} - \Gamma(0,\imath k)-\log(k)\right)\\ +\frac{1}{\imath}\left((\gamma+\imath \frac{\pi}{2})\log(k) + \frac{1}{2} (\log(k))^2 - \left.\frac{d}{d a} \Gamma(a,\imath k)\right|_{a=0} + (\log(k) + \imath \frac{\pi}{2}) \Gamma(0,\imath k)\right) \end{eqnarray} The result was obtained by expanding the exponential in the integrand in a series, then by using the former anti-derivatives (on top) and summing up each term in the series. Now the only thing we need to do is to take the anti-derivative in the limits from zero to infinity. This is pretty straightforward. The only difficulty is dealing with the $F_{3,3}$ function at infinity. Using the integral representation of that hypergeometric function we easily find that: \begin{eqnarray} x F_{3,3}\left[\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}; -x\right] \stackrel{=}{x\rightarrow \infty} \frac{1}{2!} \left[ (\log(x))^2 + 2 \gamma \log(x) + \gamma^2 + \frac{\pi^2}{6} \right] \end{eqnarray} From all this we easily arrive at the final result: \begin{eqnarray} {\mathcal I}_1(A,\mu) = -\imath \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!} \frac{(-\imath A)^{n\mu}}{n\mu} \Gamma(n \mu) -\frac{\imath}{2} (\log(A))^2 + \left(-\frac{\pi}{2} + \imath \gamma(1-\frac{1}{\mu})\right)\log(A) + \frac{\imath (-2+\mu^2) \pi^2}{24 \mu^2} + \frac{(-1+\mu)\pi}{2 \mu} \gamma - \frac{\imath (-1+\mu)^2}{2 \mu^2}\gamma^2 \end{eqnarray} Now we proceed to the second integral. We have: \begin{eqnarray} {\mathcal I}_2(A,\mu) := \int\limits_0^\infty e^{-(k A)^\mu} \cdot\left(\frac{\gamma+\Gamma(0, \imath k) + \log(\imath k)}{\imath k}\right) \cdot \mu (k A)^\mu \log(k A) dk \end{eqnarray} Again, we expand the exponential in a series and we find the anti-derivatives of each term in the expansion. Then we sum up the resulting series and we obtain the anti-derivative of the integrand. The result is as follows: \begin{eqnarray} \int e^{-(k A)^\mu} \cdot\left(\frac{\gamma+\Gamma(0, \imath k) + \log(\imath k)}{\imath k}\right) \cdot \mu (k A)^\mu \log(k A) dk = \frac{1}{\imath}\sum\limits_{i=0}^2\sum\limits_{j=0}^1 {\mathcal A}_{i,j}(k) (\log(k))^i e^{-j (A k)^\mu} +\\ (\frac{\pi}{2} +\imath \log(A)) \sum\limits_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{\mu(n+1)} \Gamma(\mu(n+1),\imath k)}{n+1}-\\ \imath \sum\limits_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{\mu(n+1)}\Gamma(\mu(n+1),\imath k)}{\mu(n+1)^2}+\\ \imath\sum\limits_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{\mu(n+1)} \left. d_a \Gamma(a,\imath k)\right|_{a=\mu(n+1)}}{n+1} \end{eqnarray} where \begin{eqnarray} {\mathcal A}_{0,0}(k)&=& \frac{2 A^{\mu } k^{\mu } \, _3F_3\left(1,1,1;2,2,2;-A^{\mu } k^{\mu }\right)}{\mu ^2}-\left(\frac{\Gamma (0,i k)}{\mu }+\frac{\gamma }{\mu }+\frac{i \pi }{2 \mu }\right) \Gamma \left(0,A^{\mu } k^{\mu }\right)+\\ &&\log (A)\left(-\frac{\Gamma \left(0,A^{\mu } k^{\mu }\right)}{\mu }-\frac{\gamma }{\mu }\right)-\log ^2(A)-\frac{\gamma \Gamma (0,i k)}{\mu }-\frac{\gamma ^2}{\mu }-\frac{i \gamma \pi }{2 \mu }\\ % {\mathcal A}_{1,0}(k)&=&-\frac{2 \Gamma \left(0,A^{\mu } k^{\mu }\right)}{\mu }-2 \log (A)-\frac{2 \gamma }{\mu }\\ {\mathcal A}_{2,0}(k)&=& -1\\ {\mathcal A}_{0,1}(k)&=& \log (A) \left(-\Gamma (0,i k)-\frac{i \pi }{2}-\gamma \right)\\ {\mathcal A}_{1,1}(k)&=& -\log (A)-\Gamma (0,i k)-\frac{i \pi }{2}-\gamma \\ {\mathcal A}_{2,1}(k)&=& -1 \end{eqnarray} Now we take the right hand side in limits from zero to infinity and we get: \begin{eqnarray} {\mathcal I}_2(A,\mu) := \\ -(\frac{\pi}{2} + \imath \log(A)) \sum\limits_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{\mu(n+1)}\Gamma((n+1)\mu)}{n+1}\\ + \imath \sum\limits_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{\mu(n+1)}\Gamma((n+1)\mu)}{\mu (n+1)^2}\\ - \imath \sum\limits_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(A/\imath)^{\mu(n+1)}\Gamma((n+1)\mu) \psi((n+1)\mu)}{(n+1)} \\ -\frac{i \gamma \log (A)}{\mu }-\frac{i \gamma ^2}{\mu ^2}-\frac{i \pi ^2}{6 \mu ^2}+\frac{i \gamma ^2}{\mu }-\frac{\gamma \pi }{2 \mu } \end{eqnarray} where $\psi$ is the di-gamma function. Putting everything together we get the final result: \begin{eqnarray} {\mathcal I}(A,\mu) = \frac{\pi}{2} \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!} A^{n\mu} \cos(\frac{\pi}{2} n \mu)\Gamma(n \mu) + \log(A) \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!} A^{n\mu} \sin(\frac{\pi}{2} n \mu)\Gamma(n \mu) + \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!} A^{n\mu} \sin(\frac{\pi}{2} n \mu)\Gamma(n \mu)\Psi(n\mu) + \frac{\pi}{2}\left(\log(A)-\gamma\right) \end{eqnarray}

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