Evaluate $\int_0^\infty x^{n+\frac12}e^{-\frac x2}\log^2x\,dx$ and $\int_0^\infty x^ne^{-\frac x2}\log^2x\,dx$

Question

Determine the closed forms of $$\mathfrak I_1=\int_0^\infty x^{n+\frac12}e^{-\frac x2}\log^2x\,dx\quad\text{and}\quad\mathfrak I_2=\int_0^\infty x^ne^{-x/2}\log^2x\,dx$$ where $s>0$ is an integer.

This problem arose while determining the Jeffreys prior of the $\chi^2$-distribution as I tried to evaluate the Fisher information matrix $$\mathcal I(k)=\int_0^\infty f(x\mid k)\left(\frac d{dk}\log f(x\mid k)\right)^2\,dx$$ where $f(x\mid k)$ is the p.d.f. of the distribution. After splitting the integral into three components, the matrix became $$\mathcal I(k)=K_1^2-K_1K_2+\frac{2^{k/2-2}}{\Gamma(k/2)}\color{red}{\int_0^\infty x^{k/2}e^{-x/2}\log^2x\,dx}$$ where $$K_1=\frac12\log2+\frac1{\psi^{(0)}(k/2)}\quad\text{and}\quad K_2=\log2+\psi^{(0)}(k/2).$$ The integrals $\mathfrak I_1$ and $\mathfrak I_2$ are obtained by setting $k=2n+1$ and $2n$ respectively in the integral in red. Plugging in some values into WolframAlpha, I found that they are of the forms \begin{alignat}2\mathfrak I_1&=\sqrt{\frac\pi2}\left(a+b\pi^2+\gamma\left(-c+2b\gamma+4b\log2\right)-(c-2b\log2)\log2\right)\tag1\\\mathfrak I_2&=2^q(r+s\pi^2-\gamma(t-6s\gamma+12s\log2)+(t+6s\log2)\log2)\tag2\end{alignat} where $b=(2n+1)!/(2^n\cdot n!)$, $\gamma$ is the Euler-Mascheroni constant and $a,c,q,r,s,t$ are positive constants.

Is there a relationship between the values of $a,c$ and $b$, and similarly between the values of $q,r,t$ and $s$? Note that I have not been able to find an expression for $s$.

Can the forms of $(1)$ and $(2)$ be proven analytically?

Answer 1

Here is an analytic evaluation for $\mathfrak I_2$. The evaluation for $\mathfrak I_1$ is similar.

Enforcing a substitution of $x \mapsto 2x$ one has \begin{align} \mathfrak I_2 &= 2^{n + 1} \int_0^\infty e^{-x} x^n \log^2 (2x) \, dx\\ &= 2^{n + 1} \log^2 2 \int_0^\infty e^{-x} x^n \, dx + 2^{n + 2} \log 2 \int_0^\infty e^{-x} x^n \log x \, dx\\ & \qquad + 2^{n + 1} \int_0^\infty e^{-x} x^n \log^2 x \, dx. \tag1 \end{align}

For the first integral: $$\int_0^\infty e^{-x} x^n \, dx = \Gamma (n + 1) = n!, \quad \text{since} \,\,n = 0,1,2, \ldots$$

For the second integral: \begin{align} \int_0^\infty e^{-x} x^n \log x \, dx &= \frac{d}{ds} \left [\int_0^\infty e^{-x} x^{n + s} \, dx \right ]_{s = 0}\\ &= \frac{d}{ds} \left [\Gamma (n + s + 1) \right ]_{s = 0}\\ &= \Gamma'(n + s + 1) \Big{|}_{s = 0}\\ &= \Gamma (n + s + 1) \psi^{(0)}(n + s + 1) \Big{|}_{s = 0}\\ &= \Gamma (n + 1) \psi^{(0)} (n + 1)\\ &= n! \, \psi^{(0)} (n + 1). \end{align}

For the third integral: \begin{align} \int_0^\infty e^{-x} x^n \log^2 x \, dx &= \frac{d^2}{ds^2} \left [\int_0^\infty e^{-x} x^{n + s} \, dx \right ]_{s = 0}\\ &= \frac{d^2}{ds^2} \Gamma (n + s + 1) \Big{|}_{s = 0}\\ &= \frac{d}{ds} \Gamma'(n + s + 1) \Big{|}_{s = 0}\\ &= \frac{d}{ds} \Gamma (n + s + 1) \psi^{(0)}(n + s + 1) \Big{|}_{s = 0}\\ &= \Gamma (n + s + 1) \left (\psi^{(0)} (n + s + 1) \right )^2 + \Gamma (n + s + 1) \psi^{(1)} (n + s + 1) \Big{|}_{s = 0}\\ &= \Gamma (n + 1) \left (\psi^{(1)} (n + 1) \right )^2 + \Gamma (n + 1) \psi^{(1)} (n + 1)\\ &= n! \left [\left (\psi^{(0)} (n + 1) \right )^2 + \psi^{(1)} (n + 1) \right ]. \end{align}

So (1) becomes $$\mathfrak I_2 = 2^{n + 1} n! \left [\log^2 2 + 2 \log 2 \psi^{(0)} (n + 1) + \left (\psi^{(0)} (n + 1) \right )^2 + \psi^{(1)} (n + 1) \right ].$$ Since $n = 0,1,2,\ldots$ the above expression in terms of the polygamma function can be further reduced to an expression containing the $n$th order harmonic number $H_n$ and the $n$th order generalised harmonic number of order two $H^{(2)}_n$. Since $$\psi^{(0)}(n + 1) = -\gamma + \sum_{k = 1}^n \frac{1}{k} = -\gamma + H_n,$$ and $$\psi^{(1)} (n + 1) = \zeta (2) - \sum_{k = 1}^n \frac{1}{k^2} = \frac{\pi^2}{6} - H^{(2)}_n,$$ then $$\mathfrak I_2 = 2^{n + 1} n! \left [\log^2 2 - 2 \gamma \log 2 + 2 \log 2 H_n + \gamma^2 - 2 \gamma H_n + H^2_n + \frac{\pi^2}{6} - H^{(2)}_n \right ],$$ valid for $n = 0,1,2, \ldots$. Here $\gamma$ denotes the Euler-Mascheroni constant.

Answer 2

For $s>-1$, I found the following: $$ \int_0^{ + \infty } {x^s e^{ - x/2} \log ^2 xdx} = 2^{s + 1} \int_0^{ + \infty } {t^s e^{ - t} \log ^2 (2t)dt} \\ = 2^{s + 1} \int_0^{ + \infty } {t^s e^{ - t} \log ^2 tdt} + 2^{s + 2} \log 2\int_0^{ + \infty } {t^s e^{ - t} \log tdt} + 2^{s + 1} \log ^2 2\int_0^{ + \infty } {t^s e^{ - t} dt} \\ = 2^{s + 1} \left[ {\frac{{d^2 }}{{dz^2 }}\int_0^{ + \infty } {t^{z + s} e^{ - t} dt} } \right]_{z = 0} + 2^{s + 2} \log 2\left[ {\frac{d}{{dz}}\int_0^{ + \infty } {t^{z + s} e^{ - t} dt} } \right]_{z = 0} + 2^{s + 1} \log ^2 2\int_0^{ + \infty } {t^s e^{ - t} dt} \\ = 2^{s + 1} \left[ {\frac{{d^2 }}{{dz^2 }}\Gamma (z + s + 1)} \right]_{z = 0} + 2^{s + 2} \log 2\left[ {\frac{d}{{dz}}\Gamma (z + s + 1)} \right]_{z = 0} + (2^{s + 1} \log ^2 2)s! \\ = 2^{s + 1} s!\left[ {s!\psi ^2 (s + 1) + \psi (s + 1)\log 4 + \psi '(s + 1) + \log ^2 2} \right]. $$

Answer 3

Both integrals we can calculate by considering the following function:

$$I(s) = \int_0^\infty x^se^{-\frac{x}{2}}\log^2(x)\:dx = \frac{d^2}{ds^2}\int_0^\infty x^se^{-\frac{x}{2}}\:dx$$

Then with an appropriate change of variables we get that

$$I(s) = \frac{d^2}{ds^2}2^{s+1}\Gamma(s+1) = 2^{s+1}\left[\log^2(2)\Gamma(s+1)+2\log(2)\Gamma'(s+1)+\Gamma''(s+1)\right]$$

$$ = 2^{s+1}\Gamma(s+1)\left[\log^2(2) + 2\log(2)\psi(s+1) + \psi^2(s+1)+\psi^{(2)}(s+1)\right]$$

where $\psi$ are the polygamma functions, with the first terms being just the digamma function. All that's left is to plug $s=n+\frac{1}{2}$ and $s=n$, respectively, for the two integrals.