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Find the asymptotic expansion of $$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt, \text{ as } x \to 0^{+}$$

I tried expanding $\frac{1}{ \sqrt{1+t^2}} = 1 - \frac{t^2}{2} + \frac{3 t^{4}}{8} + \cdot \cdot \cdot$

The integral is then :$$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt = \bigg[\frac{1}{t} - \frac{t}{2} + \frac{3t^{3}}{8} \cdot \cdot \cdot \bigg] dt$$

I observed that the first term goes to infinity as $x \to 0^{+}$.

Can someone point out if I am going in the wrong direction?

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{x}^{1}{\dd t \over t\root{1 + t^{2}}} \,\right\vert_{\ \mrm{as}\ x\ \to\ 0^{+}}} \\[5mm] = &\ \int_{x}^{1}{\dd t \over t} + \int_{x}^{1} \pars{{1 \over t\root{1 + t^{2}}} - {1 \over t}}\dd t \\[5mm] = &\ -\ln\pars{x} + \int_{x}^{1}{1 - \root{1 + t^{2}} \over t\root{1 + t^{2}}}\,\dd t \\[5mm] = &\ -\ln\pars{x} - \int_{x}^{1} {t \over 1 + t^{2} + \root{1 + t^{2}}}\,\dd t \\[5mm] = &\ -\ln\pars{x} - {1 \over 2}\int_{0}^{1} {\dd t \over 1 + t + \root{1 + t}} \\[2mm] &\ + {1 \over 2}\int_{0}^{x^{2}} {\dd t \over 1 + t + \root{1 + t}} \\[5mm] \stackrel{\mrm{as}\ x\ \to\ 0^{+}}{\sim}\,\,\, &\ -\ln\pars{x} - {1 \over 2}\int_{0}^{1} {\dd t \over 1 + t + \root{1 + t}}\,\dd t \\[2mm] & \phantom{A} + {1 \over 2}\int_{0}^{x^{2}} \pars{{1 \over 2} - {3t \over 8} + {5t^{2} \over 16}}\dd t \\[5mm] = &\ -\ln\pars{x}\ -\ \underbrace{{1 \over 2}\int_{0}^{1} {\dd t \over 1 + t + \root{1 + t}}\,\dd t} _{\mbox{a constant}} \\[2mm] &\ + {1 \over 4}\,x^{2} - {3 \over 32}\,x^{4} + {5 \over 96}\,x^{6} \end{align}

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  • $\begingroup$ What are the non-zero terms? @felix $\endgroup$ – Atul Anurag Sharma Oct 19 at 16:03
  • $\begingroup$ @AtulAnuragSharma The constant as given by an integral. $\endgroup$ – Felix Marin Oct 19 at 16:22
  • $\begingroup$ $1/4 , -3/32, 5/96$? $\endgroup$ – Atul Anurag Sharma Oct 19 at 16:24
  • $\begingroup$ Why do we still have $x$ terms in the final answer as $x \to 0^{+}?$@Felix $\endgroup$ – Atul Anurag Sharma Oct 19 at 16:25
  • $\begingroup$ @AtulAnuragSharma They appear as a result of the term a term integration. $\endgroup$ – Felix Marin Oct 19 at 16:26
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HINT

I would start by making the substitution $t = \sinh(u)$. Thus we get \begin{align*} \int\frac{\mathrm{d}t}{t\sqrt{1+t^{2}}} = \int\frac{\cosh(u)}{\sinh(u)\cosh(u)}\mathrm{d}u = \int\frac{\mathrm{d}u}{\sinh(u)} = \int\frac{\sinh(u)}{\sinh^{2}(u)}\mathrm{d}u = \int\frac{\sinh(u)}{\cosh^{2}(u)-1}\mathrm{d}u \end{align*}

Now you can make the substitution $v = \cosh(u)$, which leads to \begin{align*} \int\frac{\sinh(u)}{\cosh^{2}(u) - 1}\mathrm{d}u = \int\frac{\mathrm{d}v}{v^{2} - 1} = \frac{1}{2}\int\left(\frac{1}{v-1} - \frac{1}{v+1}\right)\mathrm{d}v \end{align*}

Can you take it from here?

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  • $\begingroup$ Now, I just need to expand $\frac{1}{v-1} \text{ and } \frac{1}{v-1}$, right? $\endgroup$ – Atul Anurag Sharma Oct 17 at 23:15

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