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When a person free-falls into a non-rotating black hole from ##r=r_0## to some position ##r=r'##, the proper time becomes

$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}+\sqrt{\frac{r'}{r_0}(1-\frac{r'}{r_0})}+\arctan(-\sqrt{\frac{r'}{r_0-r'}})].$$

Now we can simplify the above by taking the limits of ##r_0>>r'##.

First, using ##\arctan{x}=x+O(x^3)##,

$$\arctan(-\sqrt{\frac{r'}{r_0-r'}})=-\sqrt{\frac{r'}{r_0-r'}}+O[(-\sqrt{\frac{r'}{r_0-r'}})^3],$$

and here, for

$$\sqrt{\frac{r'}{r_0-r'}}=\sqrt{\frac{\frac{r'}{r_0}}{1-\frac{r'}{r_0}}} << 1,$$

we can ignore ##O(x^3)##. Then

$$\sqrt{\frac{r'}{r_0}(1-\frac{r'}{r_0})}+\arctan(-\sqrt{\frac{r'}{r_0-r'}})$$ $$≈\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}+\sqrt{\frac{r'}{r_0}}(1-\frac{r'}{r_0})^{\frac{1}{2}}-\sqrt{\frac{r'}{r_0}}(1-\frac{r'}{r_0})^{-\frac{1}{2}}].$$

By using

$$(1-\frac{r'}{r_0})^{\frac{1}{2}}≈1-\frac{r'}{r_0}$$ $$(1-\frac{r'}{r_0})^{-\frac{1}{2}}≈1+\frac{r'}{r_0},$$

I get

$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}-(\frac{r'}{r_0})^{\frac{3}{2}}].$$

This is my result, but the book says

$$τ(r')-τ(r_0)=\frac{r_0}{c}\sqrt{\frac{r_0}{r_S}}[\frac{π}{2}-\frac{2}{3}(\frac{r'}{r_0})^{\frac{3}{2}}].$$

Where did the factor ##\frac{2}{3}## come from?