I encounter an integral involving product of the modified Bessel function and Gaussian function as follows:
$$F(r)=\int_{r}^{\infty}xK_{0}(x)e^{-a^2x^2}dx,$$
where $K_{0}(x)$ is the modified Bessel function of the second kind. $a$ and $r$ are some positive constants.
I found a definite integral in Prudnikov, Anatoliĭ Platonovich, et al. Integrals and series: special functions. Vol. 2. CRC Press, 1986, shown as follows:
$$\int_0^{\infty} x e^{-p x^2} K_0(c x) d x=-\frac{1}{4 p} \exp \left(\frac{c^2}{4 p}\right) \mathrm{Ei}\left(-\frac{c^2}{4 p}\right),$$
But it starts from $0$.
I think maybe it is hard to find a close form of the integral $F(r)$, but can we get some asymptotic form of the integral in the form of, i.e.,
$$F(r)=e^{-a^2x^2}\left(K_{0}(x)\sum_{n=0}^{\infty}b_{n}x^{n}+K_{1}(x)\sum_{n=0}^{\infty}c_{n}x^{n}\right)\Bigg\vert_{x=r}^{x=\infty},$$ where the $b_{n}$ and $c_{n}$ are some recursive coefficients.
Best Answer
I shall discuss the asymptotic properites of the integral when $r$ becomes large. Using the relations $$\tag{1} K'_0 (r) = - K_1 (r),\quad K'_1 (r) = -K_0 (r) - \frac{{K_1 (r)}}{r} $$ and performing integration by parts, we find \begin{align*} \int_r^{ + \infty } {xK_0 (x){\rm e}^{ - a^2 x^2 } {\rm d}x} & = \frac{1}{{2a^2 }}{\rm e}^{ - a^2 r^2 } K_0 (r) - \frac{1}{{2a^2 }}\int_r^{ + \infty } {{\rm e}^{ - a^2 x^2 } K_1 (x){\rm d}x} \\ & = \frac{1}{{2a^2 }}{\rm e}^{ - a^2 r^2 } K_0 (r) - \frac{1}{{4a^4 }}\frac{1}{r}{\rm e}^{ - a^2 r^2 } K_1 (r) \\ & \quad\,+ \frac{1}{{4a^4 }}\int_r^{ + \infty } {\frac{1}{x}{\rm e}^{ - a^2 x^2 } K_0 (x){\rm d}x} + \frac{1}{{2a^4 }}\int_r^{ + \infty } {\frac{1}{{x^2 }}{\rm e}^{ - a^2 x^2 } K_1 (x){\rm d}x} . \end{align*} Continuing in this way, it is suspected that the integral has an asymptotic expansion of the form $$\tag{2} \int_r^{ + \infty } {xK_0 (x){\rm e}^{ - a^2 x^2 } {\rm d}x} \sim \frac{1}{{2a^2 }}{\rm e}^{ - a^2 r^2 } \left[ {K_0 (r)\sum\limits_{n = 0}^\infty {\frac{{A_n (a)}}{{r^{2n} }}} - \frac{1}{{2a^2 }}\frac{{K_1 (r)}}{r}\sum\limits_{n = 0}^\infty {\frac{{B_n (a)}}{{r^{2n} }}} } \right] $$ as $r\to +\infty$, where $A_n (a)$ and $B_n (a)$ are polynomials in $a^{-2}$. Differentiating both sides of $(2)$, using $(1)$, and equating like powers of $r$, we find that $ A_0 (a) = B_0 (a) = 1$ and $$ A_n (a) = - \frac{{n - 1}}{{a^2 }}A_{n - 1} (a) + \frac{1}{{4a^4 }}B_{n - 1} (a),\quad B_n (a) = A_n (a) - \frac{n}{{a^2 }}B_{n - 1} (a) $$ for $n\geq 1$.
To prove the asymptotic character of the expansion $(2)$, one may introduce a remainder term and show that it has the right order of magnitude. I am not going to pursue this here, for there is a simpler asymptotic expansion for the integral. Indeed, introducing the known large-$r$ asymptotic series of $K_{0,1}(r)$ in $(2)$ shows that there should be an asymptotic expansion of the form $$\tag{3} \int_r^{ + \infty } {xK_0 (x){\rm e}^{ - a^2 x^2 } {\rm d}x} \sim \frac{1}{{2a^2 }}\sqrt {\frac{\pi }{{2r}}} {\rm e}^{ - a^2 r^2 -r} \sum\limits_{n = 0}^\infty {\frac{{C_n (a)}}{{r^n }}} $$ as $r\to +\infty$, where the $C_n (a)$ are polynomials in $a^{-2}$. Differentiating both sides of $(3)$, using the asymptotic expansion of $K_0(r)$, and equating like powers of $r$, we find that $$ C_0 (a) = 1,\quad C_1 (a) = - \frac{1}{{2a^2 }} - \frac{1}{8} $$ and $$ C_n (a) = ( - 1)^n \frac{{(2n)!^2 }}{{32^n n!^3 }} - \frac{1}{{2a^2 }}C_{n - 1} (a) - \frac{{2n - 3}}{{4a^2 }}C_{n - 2} (a) $$ for $n\geq 2$. It is seen that $C_n(a)$ is a polynomial in $a^{-2}$ of degree $n$.
Finally, I shall demonstrate the asymptotic property of the expansion in $(3)$. For each $N\geq 0$ and $r \gg 0$, we write $$\tag{4} \int_r^{ + \infty } {xK_0 (x){\rm e}^{ - a^2 x^2 } {\rm d}x} = \frac{1}{{2a^2 }}\sqrt {\frac{\pi }{{2r}}} {\rm e}^{ - a^2 r^2 - r} \sum\limits_{n = 0}^{N - 1} {\frac{{C_n (a)}}{{r^n }}} + R_N (a,r) $$ We know that, for each $N\geq 0$, $$\tag{5} K_0 (r) = \sqrt {\frac{\pi }{{2r}}} {\rm e}^{ - r} \sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{{(2n)!^2 }}{{32^n n!^3 r^n }}} + \varepsilon _N (r), $$ with $$ \varepsilon _N (r) = \frac{{{\rm e}^{ - r} }}{{\sqrt r }}\mathcal{O}\!\left( {\frac{1}{{r^N }}} \right) $$ for $r\gg 0$. Differentiating both sides of $(4)$, using $(5)$, and simplifying the expression gives \begin{align*} \sqrt {\frac{{2r}}{\pi }} \frac{1}{r}{\rm e}^{a^2 r^2 + r} R'_N (a,r) = & - \sqrt {\frac{{2r}}{\pi }} {\rm e}^r \varepsilon _N (r) + \frac{1}{{2a^2 }}\frac{{C_{N - 1} (a)}}{{r^N }} + \frac{1}{{4a^2 }}\frac{{(2N - 3)C_{N - 2} (a)}}{{r^N }} \\ & + \frac{1}{{4a^2 }}\frac{{(2N - 1)C_{N - 1} (a)}}{{r^{N + 1} }} = \mathcal{O}\!\left( {\frac{1}{{r^N }}} \right) \end{align*} as $r\to +\infty$. It is clear from $(4)$ that $\lim_{r\to+\infty}R_N (a,r)=0$, whence $$ R_N (a,r) = \mathcal{O}(1)\int_r^{ + \infty } { \sqrt t {\rm e}^{ - a^2 t^2 - t} \frac{1}{{t^N }}\mathrm{d}t} $$ as $r\to +\infty$. By L'Hôpital's rule $$ \int_r^{ + \infty } {\sqrt t {\rm e}^{ - a^2 t^2 - t} \frac{1}{{t^N }}\mathrm{d}t} \sim \frac{1}{{2a^2 }}\frac{{{\rm e}^{ - a^2 r^2 - r} }}{{\sqrt r }}\frac{1}{{r^N }} $$ and therefore $$ R_N (a,r) = \frac{{{\rm e}^{ - a^2 r^2 - r} }}{{\sqrt r }}\mathcal{O}\!\left( {\frac{1}{{r^N }}} \right) $$ as $r\to +\infty$. This proves that for each $N\geq 0$, $$ \int_r^{ + \infty } {xK_0 (x){\rm e}^{ - a^2 x^2 } {\rm d}x} = \frac{1}{{2a^2 }}\sqrt {\frac{\pi }{{2r}}} {\rm e}^{ - a^2 r^2 - r}\! \left(\sum\limits_{n = 0}^{N - 1} {\frac{{C_n (a)}}{{r^n }}} +\mathcal{O}\!\left( {\frac{1}{{r^N }}} \right)\right) $$ as $r\to +\infty$. $\quad \square$