# Improper integral involving logarithm, exponent, and modified Bessel function

## Question

How can this integral, plotted in Fig. 1, be expressed more simply, for evaluation, perhaps in terms of common special functions?

$$H = \int_0^\infty \log(a)\,e^{-a^2}I_0(ba)\,da,\tag{1}$$

where $$I_0$$ is a modified Bessel function of the first kind and $$b > 0.$$

Figure 1. Plot of the integral $$H$$ in Eq. (1) as function of variable $$b$$.

## My attempts

I have not been able to find such integrals in tables of integrals and Wolfram Alpha times out. Mathematica gives:

$$H = \frac{1}{4}\, \sqrt{\pi}\Bigg(L_{-1/2}\left(\frac{b^2}{4}\right)\Psi\left(\frac{1}{2}\right) - L^{(1,\,0)}_{-1/2}\left(\frac{b^2}{4}\right)\Bigg),\tag{2}$$

where $$L_n(x)$$ is Laguerre's L function, $$\gamma$$ is the Euler–Mascheroni constant, $$\Psi$$ is the digamma function, and the superscript $$(1,0)$$ denotes taking the derivative of $$L_n(x)$$ with respect to the variable $$n$$. $$L$$ is defined in terms of the confluent hypergeometric function of the first kind, $${_1F_1}:$$

$$L_n(x) = {_1F_1}(-n;1;x) = \sum_{k=0}^\infty \frac{(-n)_k\,x^k}{(1)_k\,k!},\tag{3}$$

where $$(x)_k$$ is a Pochhammer symbol with special cases:

$$\begin{eqnarray}(1)_k &=& k!,\\ (1/2)_k &=& (2k - 1)!!/2^k,\end{eqnarray}\tag{4}$$

where $$!!$$ denotes a double factorial rather than a factorial of a factorial. The first special function in Eq. (2) has a series ($$x = b^2/4$$ to be plugged in):

$$L_{-1/2}(x) = {_1}F_1\left(\frac{1}{2};1;x\right) = \sum_{k=0}^\infty\frac{(1/2)_k\,x^k}{(1)_k\,k!}.\tag{5}$$

$${_1F_1}\left(\frac{1}{2} + \nu;\,2\nu + 1;\,x\right) = x^{-\nu-1/2}\,e^{x/2}\,4^\nu\sqrt{x}\,I_\nu\left(\frac{x}{2}\right)\,\Gamma(\nu+1).\tag{6}$$

At $$\nu = 0,$$ we get:

$$L_{-1/2}(x) = e^{x/2}\,I_0\left(\frac{x}{2}\right).\tag{7}$$

The second special function in Eq. (2), the derivative of $$L_n(x)$$ with respect to $$n$$, with $$n=-1/2,$$ does not seem to be very common. A good source of identities is (Ancarani & Gasaneo, 2008), for example their Eq. (5a) helps to get:

$$\begin{array}{l}\displaystyle L_{-1/2}^{(1, 0)}(x) = -\,{_1F_1}^{(1, 0, 0)}\left(\frac{1}{2};1;x\right)\\ \displaystyle = -\sum_{k=0}^\infty\frac{(1/2)_k}{(1)_k}\Psi\left(\frac{1}{2}+k\right)\frac{x^k}{k!} + \Psi\left(\frac{1}{2}\right)\,{_1F_1}\left(\frac{1}{2};1;x\right)\\ \displaystyle = -\sum_{k=0}^\infty\frac{(1/2)_k\,x^k}{(1)_k\,k!}\Psi\left(\frac{1}{2}+k\right)+\Psi\left(\frac{1}{2}\right)\,e^{x/2}\,I_0\left(\frac{x}{2}\right)\\ \displaystyle = -\sum_{k=0}^\infty\frac{(1/2)_k\,x^k}{(1)_k\,k!}\left(\Psi\left(\frac{1}{2}\right) + 2\sum_{n=1}^k \frac{1}{2n - 1}\right)+\Psi\left(\frac{1}{2}\right)\,e^{x/2}\,I_0\left(\frac{x}{2}\right).\end{array}\tag{8}$$

Recognizing the series of $$e^{z/2}\,I_0\left(\frac{x}{2}\right),$$ see Eqs. (5) and (7), some terms cancel and we get:

$$L_{-1/2}^{(1, 0)}(x) = -\sum_{k=0}^\infty\left(\frac{(1/2)_k\,x^k}{(1)_k\,k!}\times 2\sum_{n=1}^k \frac{1}{2n - 1}\right)\\ = 0 - x - \frac{1}{2}x^2 - \frac{161}{144}x^3 - \frac{23595}{32}x^4 - \frac{1524360005233065}{1024}x^5 - \frac{26034220296741617347072389763198341076875}{2048}x^6 - \ldots.\tag{9}$$

That doesn't seem to converge. Combining the series of Eqs. (5) and (9) into Eq. (2) gives:

$$\begin{eqnarray}H &=& \frac{\sqrt{\pi}}{4}\,\left(\Psi\left(\frac{1}{2}\right)\sum_{k=0}^\infty\frac{(1/2)_k\,\left(\frac{b^2}{4}\right)^k}{(1)_k\,k!} + \sum_{k=0}^\infty\Bigg(\frac{(1/2)_k\,\left(\frac{b^2}{4}\right)^k}{(1)_k\,k!}\times 2\sum_{n=1}^k \frac{1}{2n - 1}\Bigg)\right)\\ &=&\frac{\sqrt{\pi}}{2}\,\sum_{k=0}^\infty\left(\frac{(1/2)_k\,\left(\frac{b^2}{4}\right)^k}{(1)_k\,k!}\left(\sum_{n=1}^k \frac{1}{2n - 1} + \frac{\Psi(1/2)}{2}\right)\right),\end{eqnarray}\tag{10}$$

where $$\Psi\left(\frac{1}{2}\right) = \psi^{(0)}\left(\frac{1}{2}\right) = -\gamma - \log 4 \approx -1.96351.$$ The combined series seems to converge well, and seems identical to @user150203's series, with some of the early terms given in my edit to his answer.

## Problem background

I encountered this integral in the context of calculating a minimum mean square log-magnitude error estimator of the magnitude of a complex number from its noisy observation polluted by additive circularly symmetric complex Gaussian noise of known variance, assuming an improper uniform prior of the magnitude. From this context the full expression I want to simplify is:

$$G = \exp\left(\frac{2\int_0^\infty \log(a) e^{-a^2}I_0(2ma)da}{\sqrt{\pi} e^{m^2/2} I_0(m^2/2)}\right).\tag{11}$$

$$G$$ approaches $$m$$ asymptotically as $$m\to\infty$$ (Fig. 2).

Figure 2. the desired estimator $$G(m).$$

## References

Ancarani L. U. and Gasaneo G., Derivatives of any order of the confluent hypergeometric function $${_1F_1}(a,b,z)$$ with respect to the parameter $$a$$ or $$b,$$ Journal of Mathematical Physics 49, 063508 (2008); https://doi.org/10.1063/1.2939395

• I am afraid that there is probably no closed form even for $b=1$. In any manner, it seems to vary very fast with $b$. – Claude Leibovici Feb 26 '19 at 11:31

NOT A CLOSED FORM SOLUTION:

We first observe: $$$$H = \int_0^\infty \log(a)\:e^{-a^2} I_0\left(ba\right)da = \lim_{t\rightarrow 0^+} \frac{\partial}{\partial t} \underbrace{\int_0^\infty a^t\:e^{-a^2} I_0\left(ba\right)da}_{J(t,b)}$$$$

We now employ the following definition for the Bessel Function: $$$$I_\alpha(x) = \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}$$$$

And thus: $$$$I_0(ba) = \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{ba}{2}\right)^{2m} = \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m}a^{2m}$$$$

We now address $$J(t,b)$$: \begin{align} J(t,b) &= \int_0^\infty a^t\:e^{-a^2} I_0\left(ba\right)da = \int_0^\infty a^t\:e^{-a^2} \left[ \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m}a^{2m}\right]da \\ &= \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \int_0^\infty a^t\:e^{-a^2}a^{2m}\:da = \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \int_0^\infty a^{t + 2m}\:e^{-a^2}\:da \nonumber \end{align} Making the substitution $$u = a^2$$: \begin{align} J(t,b) &= \sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \int_0^\infty \left(\sqrt{u}\right)^{t + 2m}\:e^{-u}\frac{1}{2\sqrt{u}}\:du \nonumber \\ &= \frac{1}{2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \int_0^\infty u^{\frac{t}{2} + m - \frac{1}{2}}\:e^{-u}\:du\nonumber \nonumber \\ &= \frac{1}{2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \Gamma\left(\frac{t}{2} + m + \frac{1}{2} \right) \end{align} We may now return to $$H$$: \begin{align} H &= \lim_{t\rightarrow 0^+} \frac{\partial}{\partial t} \int_0^\infty a^t\:e^{-a^2} I_0\left(ba\right)da = \lim_{t\rightarrow 0^+} \frac{\partial}{\partial t} \left[\frac{1}{2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \Gamma\left(\frac{t}{2} + m + \frac{1}{2} \right) \right] \nonumber \\ &=\lim_{t\rightarrow 0^+} \frac{1}{2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \Gamma\left(\frac{t}{2} + m + \frac{1}{2} \right) \psi^{(0)}\left(\frac{t}{2} + m + \frac{1}{2} \right)\cdot \frac{1}{2} \nonumber \\ &=\frac{1}{2^2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \Gamma\left(m + \frac{1}{2} \right) \psi^{(0)}\left(m + \frac{1}{2} \right) \nonumber \end{align}

At this moment, I'm unsure how to proceed to achieve a closed form. I hope this can serve some value in achieving that.

Actually maybe I can go a bit further. We note that $$$$\Gamma\left(m + \frac{1}{2}\right) = {m - \frac{1}{2} \choose m} m! \sqrt{\pi}$$$$

Thus, \begin{align} H &= \frac{1}{2^2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} \Gamma\left(m + \frac{1}{2} \right) \psi^{(0)}\left(m + \frac{1}{2} \right)\cdot \nonumber \\ &= \frac{1}{2^2}\sum_{m=0}^\infty \frac{1}{m!\, \Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} {m - \frac{1}{2} \choose m} m! \sqrt{\pi} \cdot \psi^{(0)}\left(m + \frac{1}{2} \right) \nonumber \\ &=\frac{\sqrt{\pi}}{2^2}\sum_{m=0}^\infty \frac{1}{\Gamma(m+1)}\left(\frac{b}{2}\right)^{2m} {m - \frac{1}{2} \choose m} \psi^{(0)}\left(m + \frac{1}{2} \right) \nonumber \\ &=\frac{\sqrt{\pi}}{2^2}\sum_{m=0}^\infty \frac{1}{m!}\left(\frac{b}{2}\right)^{2m} {m - \frac{1}{2} \choose m} \psi^{(0)}\left(m + \frac{1}{2} \right) \nonumber \end{align}

MAYBE NOT

@OlliNiemitalo's edit: The last series seems to converge well. A truncated sum of a first few terms is:

$$H \approx \sqrt{\pi}\left( \frac{\Psi\left(\frac{1}{2}\right)}{4} + \frac{2 + \Psi\left(\frac{1}{2}\right)}{32}b^2 + \frac{8 + 3\Psi\left(\frac{1}{2}\right)}{1024}b^4 + \frac{46 + 15\Psi\left(\frac{1}{2}\right)}{73728}b^6 + \frac{352 + 105\Psi\left(\frac{1}{2}\right)}{9437184}b^8 + \frac{1126 + 315\Psi\left(\frac{1}{2}\right)}{629145600}b^{10} + \frac{13016 + 3465\Psi\left(\frac{1}{2}\right)}{181193932800}b^{12} + \frac{176138 + 45045\Psi\left(\frac{1}{2}\right)}{71028021657600}b^{14} + \frac{182144 + 45045\Psi\left(\frac{1}{2}\right)}{2424423139246080}b^{16}+ \frac{3186538 + 765765\Psi\left(\frac{1}{2}\right)}{1571026194231459840}b^{18} + \frac{62075752 + 14549535\Psi\left(\frac{1}{2}\right)}{1256820955385167872000}b^{20} \right)\\ \approx - 0.87005772672831550673 + 0.0020211498405553133642\,b^{2} + 0.0036513067251018966186\,b^{4} + 0.00039780563066887487591\,b^{6} + 2.7389474475950572625\cdot 10^{-5}\,b^{8} + 1.4297331181239024928\cdot 10^{-6}\,b^{10} + 6.0770573610132180363\cdot 10^{-8}\,b^{12} + 2.1882838851707893018\cdot 10^{-9}\,b^{14} + 6.8500762234545623886\cdot 10^{-11}\,b^{16} + 1.8987287033642213687\cdot 10^{-12}\,b^{18} + 4.7254673816159087164\cdot 10^{-14}\,b^{20},$$

where $$\Psi\left(\frac{1}{2}\right) = \psi^{(0)}\left(\frac{1}{2}\right) = -\gamma - \log 4 \approx -1.96351.$$ The integer sequences in the numerators seem identical to OEIS A074599 and A025547 (see also A001147), but I can't find a sequence related to the denominator integer sequences in OEIS.

• No worries mate. – user150203 Mar 5 '19 at 12:17
• @OlliNiemitalo - I'm glad it serves something of use. – user150203 Mar 6 '19 at 3:11

Another representation as a series:

The product $$I_0(z)\ln z$$ can be found in the series expansion of $$K_0(z)$$ $$$$K_{0}\left(z\right)=-\left(\ln\left(\tfrac{1}{2}z\right)+\gamma\right)I_{0} \left(z\right)+\frac{\tfrac{1}{4}z^{2}}{(1!)^{2}}+(1+\tfrac{1}{2})\frac{( \tfrac{1}{4}z^{2})^{2}}{(2!)^{2}}+(1+\tfrac{1}{2}+\tfrac{1}{3})\frac{(\tfrac{1 }{4}z^{2})^{3}}{(3!)^{2}}+\cdots$$$$ and thus $$$$\ln\left(a\right)I_{0}\left(ab\right)=- K_{0}\left(ab\right)+\left(\gamma +\ln \frac{2}{b} \right) I_{0}\left(ab\right)+\sum_{k=1}^\infty\frac{2^{-2k}H_{k}\left( ab \right)^{2k}}{(k!)^{2}}$$$$ where $$H_k=\psi(k+1)+\gamma$$ are the harmonic numbers.Using the tabulated integrals for $$I_0$$ and $$K_0$$ \begin{align} H&=\int_0^\infty \ln\left(a\right)I_{0}\left(ab\right)e^{-a^2}\,da\\ &=- \int_0^\infty K_{0}\left(ab\right)e^{-a^2}\,da-\left(\gamma +\ln \frac b2 \right) \int_0^\infty I_{0} \left(ab\right)e^{-a^2}\,da+\sum_{k=1}^\infty\frac{2^{-2k}b^{2k}H_{k}}{(k!)^{2}}\int_0^\infty a^{2k}e^{-a^2}\,da\\ &=-\frac{\sqrt{\pi}}{4}e^{\frac{b^2}{8}}\left[2\left(\gamma +\ln \frac b2 \right)I_0\left( \frac{b^2}{8} \right)+K_0\left( \frac{b^2}{8} \right)\right]+\sum_{k=1}^\infty2^{-2k-1}H_{k}\Gamma\left( k+\tfrac{1}{2} \right)\frac{b^{2k}}{(k!)^{2}} \end{align} where the series converges, as $$H_k\sim\ln k$$ for $$k\to\infty$$.

An equivalent expression can be found by using the $$\psi$$ function and identifying the term (from the hypergeometric representation) $$$$\gamma\sum_{k=1}^\infty2^{-2k-1}\Gamma\left( k+\tfrac{1}{2} \right)\frac{b^{2k}}{(k!)^{2}}=\frac{\gamma\sqrt{\pi}}{2}\left[e^{ \tfrac{b^2}{8}}I_0\left( \frac{b^2}{8} \right)-1\right]$$$$ and thus $$$$H=-\frac{\gamma\sqrt{\pi}}{2}-\frac{\sqrt{\pi}}{4}e^{\frac{b^2}{8}}\left[2\ln \frac b2 I_0\left( \frac{b^2}{8} \right)+K_0\left( \frac{b^2}{8} \right)\right]+\sum_{k=1}^\infty2^{-2k-1}\psi(k+1)\Gamma\left( k+\tfrac{1}{2} \right)\frac{b^{2k}}{(k!)^{2}}$$$$

Using the question's Eqs. (2, 7 & 8) and $$\Psi\left(\frac{1}{2}\right)$$ $$=$$ $$-\gamma - \log 4$$ we get:

$$H = \frac{\sqrt{\pi}}{4}\Bigg(e^{b^2/8}\,I_0\left(\frac{b^2}{8}\right) (-\gamma - \log 4) +\,{_1F_1}^{(1, 0, 0)}\left(\frac{1}{2};1;\frac{b^2}{4}\right)\Bigg)\tag{1}$$

There is an identity (functions.wolfram.com and Ancarani & Gasaneo, 2008 Eq. (17a) referenced in the question):

$$\begin{eqnarray}&&{_1F_1}^{(1, 0, 0)}(a;b;x)\\ &=& \frac{x}{b}F^{1\,1\,2}_{2\,0\,1}\left(\begin{array}{c}a+1;\,1;\,1,a;\\2,\,b+1;;a+1;\end{array}x,x\right)\\ &=& \frac{x}{b}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(1)_m\,(1)_n\,(a)_m\,(a+1)_{m+n}}{(a+1)_m\,(2)_{m+n}\,(b+1)_{m+n}}\frac{x^{m+n}}{m!\,n!}.\end{eqnarray}\tag{2}$$

The special function $$F^{1\,1\,2}_{2\,0\,1}$$ is a Kampé de Fériet (-like) function. It is available for example in Python's mpmath library as hyper2d. For Eq. (1) we need:

$$\,{_1F_1}^{(1, 0, 0)}\left(\frac{1}{2};1;\frac{b^2}{4}\right) = \frac{b^2}{4}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(1)_m\,(1)_n\,(1/2)_m\,(3/2)_{m+n}}{(3/2)_m\,(2)_{m+n}\,(2)_{m+n}}\frac{(b^2/4)^{m+n}}{m!\,n!},\tag{3}$$

which in Python is expressed by:

mpf(b)**2/4*hyper2d({'m+n':[1.5], 'n':[1], 'm':[1, 0.5]}, {'m+n':[2, 2], 'm':[1.5]}, mpf(b)**2/4, mpf(b)**2/4)


The full $$H:$$

$$H = \frac{\sqrt{\pi}}{4}\Bigg(e^{b^2/8}\,I_0\left(\frac{b^2}{8}\right) (-\gamma - \log 4) + \frac{b^2}{4}F^{1\,1\,2}_{2\,0\,1}\left(\begin{array}{c}3/2;\,1;\,1,1/2;\\2,\,2;;3/2;\end{array}\,\frac{b^2}{4},\frac{b^2}{4}\right)\Bigg),\tag{4}$$

is then expressed in Python by:

sqrt(mp.pi)/4*(exp(mpf(b)**2/8)*besseli(0, mpf(b)**2/8)*(-log(mpf(4))-mp.euler) + mpf(b)**2/4*hyper2d({'m+n':[1.5], 'n':[1], 'm':[1, 0.5]}, {'m+n':[2, 2], 'm':[1.5]}, mpf(b)**2/4, mpf(b)**2/4))


which evaluates correctly. I will accept this answer on the basis of the common availability of all special functions in Eq. (4). I also appreciate a lot @user150203's and @PaulEnta's series representations.

• You can express ${}_1F_1(1/2;1;b^2/4)=\exp(b^2/8)I_0(b^2/8)$ (see for example here). – Paul Enta Mar 7 '19 at 13:06
• @PaulEnta Thanks, incorporated. – Olli Niemitalo Mar 7 '19 at 13:48