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When I am reading Brian Hall's "Quantum Theory for Mathematicians", I came across an integration (frequently appeared in physics textbooks) $$t=\int_{x_0}^{x_1}\sqrt{\frac{m}{2(E_0-V(y))}}dy.$$ The potential function V is smooth, and $V(x)<E_0$ for any $x\in[x_0,x_1]$. However, $V(x_1)=E_0$.

The book says if $V'(x_1)\neq 0$, then the above integral is convergent, whereas if $V'(x_1)=0$, it is divergent.

I wonder how this can be proved. I checked Zorich's "mathematical analysis" but it just says it's obvious. So I'm sorry if it is really obvious and I just fails to see it.

Thanks in advance for any help or remark.

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1 Answer 1

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Hint. A potential problem for convergence is near $x_1$. Since $x \mapsto V(x)$ is smooth, then as $x \to x_1^-$, by the Taylor expansion we have $$ V(x)=V(x_1)+(x-x_1)V'(x_1)+\mathcal{O}\left( x-x_1\right)^2 $$ or $$ V(x)=E_0-(x_1-x)V'(x_1)+\mathcal{O}\left( x_1-x\right)^2. \tag1 $$

Case 1. $V'(x_1)\neq0.$ Clearly, since $V(x)<E_0$ for any $x\in[x_0,x_1]$, then $V'(x_1)>0$ and from identity $(1)$ we may write, as $x \to x_1^-$, $$ \sqrt{E_0-V(x)}=\sqrt{V'(x_1)}(x_1-x)^{1/2}\left(1+ \mathcal{O}\left( x_1-x\right)\right) \tag2 $$ then, for some $\alpha$ sufficiently near $x_1$,

$$ \int_{\alpha}^{x_1}\sqrt{\frac{m}{2(E_0-V(x))}}dx \sim \sqrt{\frac{m}{2V'(x_1)}}\int_{\alpha}^{x_1}\frac1{(x_1-x)^{1/2}}dx \tag3 $$

the latter integral is convergent, thus in this case your initial integral is convergent.

Case 2. $V'(x_1)=0.$ Excluding the case where $x \mapsto V(x)$ is a constant function, we then have by the Taylor expansion, for some $p\geq 2$ and for some constant $C$, as $x \to x_1^-$, $$ \sqrt{E_0-V(x)}=C(x_1-x)^{p/2}\left(1+ \mathcal{O}\left( x_1-x\right)\right) \tag4$$ then, for some $\alpha$ sufficiently near $x_1$,

$$ \int_{\alpha}^{x_1}\sqrt{\frac{m}{2(E_0-V(x))}}dx \sim \frac1C\sqrt{\frac{m}2}\int_{\alpha}^{x_1}\frac1{(x_1-x)^{p/2}}dx \tag5$$

the latter integral is divergent ($p/2\geq1$), thus in this case your initial integral is divergent.

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