# Asymptotic expansion of an integral as $x \to 0^{+}$

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?


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

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?

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