$\lim_{z\to \infty} \ln \frac{z+\sqrt{z^2+p^2}}{-z+\sqrt{z^2+p^2}}$

Question

The limit is the result of this integral: $$\int _{-\infty}^{\infty} \frac{dz}{\sqrt{z^2+p^2}}$$ This is a part from a physics problem and the result should be $\ln{p}$. I really try to do the limit by different ways and in all of them I get different results: $$\lim_{z\to \infty} \ln(\frac{z+\sqrt{z^2+p^2}}{-z+\sqrt{z^2+p^2}})=\lim_{z\to \infty} \ln(\frac{z+\sqrt{z^2+p^2}}{-z+\sqrt{z^2+p^2}}\times \frac{z-\sqrt{z^2+p^2}}{z-\sqrt{z^2+p^2}})$$

$$=\lim_{z\to \infty} \ln(\frac{(z+\sqrt{z^2+p^2})(z-\sqrt{z^2+p^2})}{(-z+\sqrt{z^2+p^2})(z-\sqrt{z^2+p^2})})=\lim_{z\to \infty} \ln(\frac{-p^2}{(-z+\sqrt{z^2+p^2})(z-\sqrt{z^2+p^2})})$$ $$=\lim_{z\to \infty} \ln(\frac{p^2}{(-z+\sqrt{z^2+p^2})(-z+\sqrt{z^2+p^2})})=\lim_{z\to \infty} \ln(\frac{p^2}{2z^2+p^2-2z\sqrt{z^2+p^2}})$$ But $-2z\sqrt{z^2+p^2}=-2z^2$ (because $z \to \infty$), so: $$==\lim_{z\to \infty} \ln(\frac{p^2}{p^2})=0$$ Of course, I made a mistake since I didn't get the correct answer $\ln(p)$. Please if someone knows the magic trick, help me.

Answer 1

To begin with, I would recommend noticing the integrand is even. Hence it follows that \begin{align*} \int_{-\infty}^{+\infty}\frac{\mathrm{d}z}{\sqrt{z^{2} + p^{2}}} = 2\int_{0}^{\infty}\frac{\mathrm{d}z}{\sqrt{z^{2} + p^{2}}} \end{align*}

On the other hand, according to the substitution $z = p\sinh(w)$, we get that \begin{align*} \int_{0}^{\infty}\frac{\mathrm{d}z}{\sqrt{z^{2} + p^{2}}} = \int_{0}^{\infty}\frac{p\cosh(w)}{\sqrt{p^{2}\sinh^{2}(w) + p^{2}}}\mathrm{d}w = \int_{0}^{\infty}\mathrm{d}w = +\infty \end{align*}

and we conclude the proposed integral diverges, as observed by @David.

Hopefully this helps!

Answer 2

Something seems to be off. In your logarithm, your numerator goes to infinity and your denominator goes to zero, so it doesn't go to $\ln p$, it goes to infinity. Similarly, in your original integral, your integrand is $\sim 1/z$ for large $z$. That integrates to $\ln z$, which diverges at infinity. So your integral doesn't converge either.