Given:
$T=l \sqrt{\frac{2m} {E}}(\frac{E} {V_0} )^t \int_{0 }^{1} ds(1-s^{\frac{1 } {t}})^{-\frac{1 } {2 }} $
$l,t, V_0>0$.
Need to calculate the time period $(T)$ when $t \to 0$
So I tried to solve by two different ways but got different solutions. Which one is the correct and why?
first way:
$\lim_{t\to 0} l\sqrt{\frac{2m} {E}}(\frac{E} {V_0} )^t \int_{0 }^{1} ds(1-s^{\frac{1 } {t}})^{-\frac{1 } {2 }} $
$= l\sqrt{\frac{2m} {E}} \lim_{t\to 0} \int_{0 }^{1} ds(-s^{-\frac{1 } {2t}})=-l\sqrt{\frac{2m} {E} } \lim_{t\to 0} \frac{1} {1-\frac{1}{2t}} s^{1-\frac{1 } {2 t} } \big|_0^1=0$
second way:
$ l\sqrt{\frac{2m} {E}}(\frac{E} {V_0} )^t \int_{0 }^{1} ds(1-s^{\frac{1 } {t}})^{-\frac{1 } {2 }}=l\sqrt{\frac{2m} {E}}(\frac{E} {V_0} ) \int_{0 }^{1} dk(1-k) ^{-\frac{1 } {2}}tk^{t-1}=l\sqrt{\frac{2m} {E}}(\frac{E} {V_0} )^t t \beta (t, 0.5)$
The substitution I did: $k=s^{\frac{1 } {t} } $
Eventually I got :
$\lim_{t\to 0} l\sqrt{\frac{2m} {E}}(\frac{E} {V_0} )^t t \beta (t, 0.5)=l\sqrt{\frac{2m} {E}}$
So as you see I got 2 different answers.
Would appreciate for your insights.
Thank you
I could just state that the second method is correct, but when you call yourself @user5713492 you are hardly claiming to be the voice of authority, so let's see how we could actually make the first method fly by estimating the value of the integral. Choose any $0<\epsilon\le1/2$ and let $$I(t)=\int_0^1\left(1-s^{1/t}\right)^{-1/2}ds=\int_0^{1-\epsilon^2/16}\left(1-s^{1/t}\right)^{-1/2}ds+\int_{1-\epsilon^2/16}^1\left(1-s^{1/t}\right)^{-1/2}ds$$ Now, $$\begin{align}I_1&=\int_0^{1-\epsilon^2/16}\left(1-s^{1/t}\right)^{-1/2}ds=\int_0^{1-\epsilon^2/16}1ds+\int_0^{1-\epsilon^2/16}\left[\left(1-s^{1/t}\right)^{-1/2}-1\right]ds\\ &=1-\frac{\epsilon^2}{16}+\int_0^{1-\epsilon^2/16}\frac{1-\left(1-s^{1/t}\right)^{1/2}}{\left(1-s^{1/t}\right)^{1/2}}ds\\ &=1-\frac{\epsilon^2}{16}+\int_0^{1-\epsilon^2/16}\frac{s^{1/t}}{\left(1+\left(1-s^{1/t}\right)^{1/2}\right)\left(1-s^{1/t}\right)^{1/2}}ds\end{align}$$ If we constrain $$0<t<\frac{\ln(1-\epsilon^2/16)}{\ln(\epsilon/2)}$$ Then since $0\le s\le1-\epsilon^2/16<1$, $$0\le s^{1/t}\le(1-\epsilon^2/16)^{1/t}\le e^{\frac{\ln(\epsilon/2)}{\ln(1-\epsilon^2/16)}\ln(1-\epsilon^2/16)}=\frac{\epsilon}2<\frac12$$ So we can say that $0<\left(1-s^{1/t}\right)^{1/2}<\left(1-\frac12\right)^{1/2}=\frac1{\sqrt2}$ so $$1-\frac{\epsilon^2}{16}<I_1<1-\frac{\epsilon^2}{16}+\int_0^{1-\epsilon^2/16}\frac{\frac{\epsilon}2}{\left(1+\frac1{\sqrt2}\right)\frac1{\sqrt2}}ds<1+\frac{\epsilon}2$$ If $t<1$ then if $1-\epsilon^2/16\le s\le1$, $0<s^{1/t}\le s\le1$ so $0\le1-s\le1-s^{1/t}<1$ and $$\begin{align}\int_{1-\epsilon^2/16}^11ds&=\frac{\epsilon^2}{16}\\ &<I_2=\int_{1-\epsilon^2/16}^1\left(1-s^{1/t}\right)^{-1/2}ds\\ &\le\int_{1-\epsilon^2/16}^1(1-s)^{-1/2}ds=\left.-2(1-s)^{1/2}\right|_{1-\epsilon^2/16}^1=\frac{\epsilon}2\end{align}$$ Adding up, we have $$1=1-\frac{\epsilon^2}{16}+\frac{\epsilon^2}{16}<I(t)=I_1+I_2<1+\frac{\epsilon}2+\frac{\epsilon}2=1+\epsilon$$ That is to say $$\left|\int_0^1\left(1-s^{1/t}\right)^{-1/2}ds-1\right|<\epsilon$$ Provided $$0<t<\delta=\min\left(1,\frac{\ln\left(1-\epsilon^2/16\right)}{\ln(\epsilon/2)}\right)$$ And that completes our $\epsilon,\delta$ proof that $$\lim_{t\rightarrow0^+}\int_0^1\left(1-s^{1/t}\right)^{-1/2}ds=1$$