Asymptotic Relation: Corrections Vanish as x → ∞

[wimma]

Homework Statement

Show that the corrections to the asymptotic relation $$\int ^x _B dt \sqrt{E-V(t)} \sim x\sqrt {E} + I (x \rightarrow \infty)$$ where $$I = \int ^\infty _B dt[\sqrt{E-V(t)}-\sqrt{E}] - B\sqrt{E}$$ vanish as $$x \rightarrow \infty$$ if $$V(x) \rightarrow 0$$ faster than $$1/x$$

Homework Equations

None

The Attempt at a Solution

Don't know how to do it at all.

 

[mighty2000]

Hello,

To show that the corrections to the asymptotic relation vanish as x approaches infinity, we can use the limit definition of a function.

First, let's define the function I(x) as:

I(x) = \int ^\infty _B dt[\sqrt{E-V(t)}-\sqrt{E}] - B\sqrt{E}

Now, we can rewrite the asymptotic relation as:

\int ^x _B dt \sqrt{E-V(t)} = x\sqrt {E} + I(x)

Next, we can take the limit as x approaches infinity on both sides of the equation:

lim_{x \to \infty} \int ^x _B dt \sqrt{E-V(t)} = lim_{x \to \infty} x\sqrt {E} + lim_{x \to \infty} I(x)

Since we know that lim_{x \to \infty} x\sqrt {E} = \infty, we can rewrite the equation as:

lim_{x \to \infty} \int ^x _B dt \sqrt{E-V(t)} = \infty + lim_{x \to \infty} I(x)

Now, we need to show that lim_{x \to \infty} I(x) = 0 in order to prove that the corrections to the asymptotic relation vanish as x approaches infinity.

To do this, we can use the limit comparison test. Since we know that V(x) approaches 0 faster than 1/x, we can choose a function g(x) = 1/x that is greater than V(x) for all x greater than some value M.

Therefore, we can rewrite the integral I(x) as:

I(x) = \int ^\infty _B dt[\sqrt{E-V(t)}-\sqrt{E}] - B\sqrt{E} \leq \int ^\infty _B dt[\sqrt{E-1/x}-\sqrt{E}] - B\sqrt{E}

By comparing the two integrals, we can see that I(x) is bounded above by a convergent integral, which means that I(x) must also converge.

Therefore, we can conclude that lim_{x \to \infty} I(x) = 0, which proves that the corrections to the asympt