Finding an asymptotic expansion for $\displaystyle \int_0^\infty t \exp(-(t-xt^{-1})^2) \ dt$ as $x\to0$.

Question

I am looking to find the first few terms in an asymptotic expansion for the integral $$ \int_{0}^{\infty}t \exp\left(-\left[t-xt^{-1}\right]^{2}\right)\,\mathrm{d}t \quad\mbox{as}\quad x \to 0 $$ I am also told that Euler's constant $\gamma$ should come in useful. However, I cannot see why this is so ( Laplace's method doesn't give an expression involving $\gamma$ so I'm not sure what method to use ). I am fairly new to asymptotics, so I would appreciate a model answer that would help further my understanding of the topic. Much appreciate it.

Answer 1

Assuming $x>0$, given by a CAS, $$f(x)=\int_0^\infty t\, e^{-\left(t-\frac{x}{t}\right)^2}\,dt=x \,e^{2 x}\, K_1(2 x)$$ where appears the modified Bessel function of the second kind.

Using the expansion of the whole term, you should get $$f(x)=\frac{1}{2}+x+x^2 \left(\log (x)+\gamma +\frac{1}{2}\right)+O\left(x^3\right)$$

To check, let $x=10^{-k}$ and the results $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.587746305719 & 0.583338603719 \\ 2 & 0.509647204548 & 0.509638700563 \\ 3 & 0.500994169460 & 0.500994156449 \\ 4 & 0.500099918669 & 0.500099918651 \\ 5 & 0.500009998956 & 0.500009998956 \\ 6 & 0.500000999987 & 0.500000999987 \\ 7 & 0.500000100000 & 0.500000100000 \\ 8 & 0.500000010000 & 0.500000010000 \\ 9 & 0.500000001000 & 0.500000001000 \end{array} \right)$$

Edit

A better approximation would be $$f(x)=e^{2x}\left(\frac{1}{2}+x^2 \left(\log (x)+\gamma -\frac{1}{2}\right)+O\left(x^3\right) \right)$$ as shown below for the same table $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.583520655506 & 0.583338603719 \\ 2 & 0.509638727486 & 0.509638700563 \\ 3 & 0.500994156453 & 0.500994156449 \\ 4 & 0.500099918651 & 0.500099918651 \end{array} \right)$$

Answer 2

Asymptotic Expansion

Let $2u=t-1/t$, then $t=u+\sqrt{u^2+1}$ and $tt'=2u+\frac{2u^2+1}{\sqrt{u^2+1}}$. $$ \begin{align} f(x) &=\int_0^\infty te^{-(t-x/t)^2}\mathrm{d}t\\ &=x\int_0^\infty te^{-x(t-1/t)^2}\mathrm{d}t\\ &=x\int_{-\infty}^\infty tt'e^{-4xu^2}\mathrm{d}u\\ &=2x\int_0^\infty\frac{2u^2+1}{\sqrt{u^2+1}}e^{-4xu^2}\mathrm{d}u\\ &=\int_0^\infty\frac{u^2+2x}{\sqrt{u^2+4x}}e^{-u^2}\mathrm{d}u\tag1 \end{align} $$ Thus, $$ \begin{align} f(0) &=\int_0^\infty ue^{-u^2}\mathrm{d}u\\[3pt] &=\frac12\tag2 \end{align} $$ Starting with $(1)$, we can compute $$ \begin{align} f'(x) &=\int_0^\infty\frac{4x}{\left(u^2+4x\right)^{3/2}}e^{-u^2}\mathrm{d}u\\ &=\int_0^\infty\frac1{\left(u^2+1\right)^{3/2}}e^{-4xu^2}\mathrm{d}u\tag3 \end{align} $$ Therefore, $$ \begin{align} f'(0) &=\int_0^\infty\frac1{\left(u^2+1\right)^{3/2}}\mathrm{d}u\\[6pt] &=1\tag4 \end{align} $$ Similarly, starting with $(3)$, we can compute $$ \begin{align} f''(x) &=\int_0^\infty\frac{-4u^2}{\left(u^2+1\right)^{3/2}}e^{-4xu^2}\mathrm{d}u\\ &=\int_0^\infty\frac{-2\sqrt{u}}{\left(u+1\right)^{3/2}}e^{-4xu}\mathrm{d}u\\ &=\color{#C00}{-2\int_0^\infty\frac{e^{-4xu}}{u+1}\,\mathrm{d}u} +\color{#090}{2\int_0^\infty\frac{\sqrt{u+1}-\sqrt{u}}{\left(u+1\right)^{3/2}}e^{-4xu}\mathrm{d}u}\\[6pt] &=\color{#C00}{2(\log(x)+\gamma+\log(4))}+\color{#090}{2(2-\log(4))}+O(x\log(x))\\[12pt] &=2\log(x)+2\gamma+4+O(x\log(x))\tag5 \end{align} $$ Combining $(2)$, $(4)$, and $(5)$, we get $$ \begin{align} f(x) &\sim\overbrace{\ \ \ \ \ \ \tfrac12\ \ \ \ \ \ }^{(2)}+\overbrace{\ \ \ \ \ \ x\vphantom{\tfrac12}\ \ \ \ \ \ }^{(4)}+\overbrace{x^2\log(x)-\tfrac32x^2}^{2\log(x)}+\overbrace{(\gamma+2)x^2\vphantom{\tfrac12}}^{2\gamma+4}+O\!\left(x^3\log(x)\right)\\[6pt] &=\bbox[5px,border:2px solid #C0A000]{\tfrac12+x+x^2\log(x)+\left(\gamma+\tfrac12\right)x^2+O\!\left(x^3\log(x)\right)}\tag6 \end{align} $$

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Red Integral $$ \begin{align} \int_x^\infty\frac{e^{-u}}u\,\mathrm{d}u &=\int_x^\infty e^{-u}\,\mathrm{d}\log(u)\\ &=-\log(x)\,e^{-x}+\int_x^\infty\log(u)\,e^{-u}\,\mathrm{d}u\\ &=-\log(x)\,e^{-x}-\gamma-\int_0^x\log(u)\,e^{-u}\,\mathrm{d}u\\ &=-\log(x)\,e^{-x}-\gamma-\int_0^x\log(u)\,\mathrm{d}\left(1-e^{-u}\right)\\ &=-\log(x)\,e^{-x}-\gamma-\log(x)\left(1-e^{-x}\right)+\int_0^x\frac{1-e^{-u}}u\,\mathrm{d}u\\ &=-\log(x)-\gamma-\sum_{k=1}^\infty\frac{(-x)^k}{k\,k!}\tag7 \end{align} $$ Therefore, $$ \begin{align} \int_0^\infty\frac{e^{-4xu}}{1+u}\,\mathrm{d}u &=e^x\int_1^\infty\frac{e^{-4xu}}{u}\,\mathrm{d}u\\ &=e^{4x}\int_{4x}^\infty\frac{e^{-u}}u\,\mathrm{d}u\\ &=-e^{4x}\left(\log(4x)+\gamma+\sum_{k=1}^\infty\frac{(-4x)^k}{k\,k!}\right)\\[6pt] &=-(\log(x)+\gamma+\log(4))+O(x\log(x))\tag8 \end{align} $$

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Green Integral

Let $u=\frac{v}{1-v}$ and $v=w^2$, then $$ \begin{align} \int_0^\infty\frac{\sqrt{u+1}-\sqrt{u}}{\left(u+1\right)^{3/2}}\mathrm{d}u &=\int_0^1\frac{\sqrt{\frac1{1-v}}-\sqrt{\frac{v}{1-v}}}{\left(\frac1{1-v}\right)^{3/2}}\frac{\mathrm{d}v}{(1-v)^2}\\ &=\int_0^1\frac{1-\sqrt{v}}{1-v}\,\mathrm{d}v\\[6pt] &=\int_0^1\frac{2w}{1+w}\,\mathrm{d}w\\[12pt] &=2-\log(4)\tag9 \end{align} $$