What is $\lim_{p \to 0} \left(\int_0^1 (1+x)^p \, dx\right)^{1/p}$?

Question

What is $$\lim_{p \to 0} \left(\int_0^1 (1+x)^p \, dx\right)^{1/p}\text{ ?}$$

I used binomial to get value as $2^p-1$ so limit becomes $$(2^p-1)^{1/p}.$$ But I can't go any further.

Answer 1

Consider $$ \int_0^1(1+x)^p\,dx=\left[\frac{(1+x)^{p+1}}{p+1}\right]_0^1= \frac{2^p-1}{p+1} $$ (we may consider $p>-1$, of course). Now you want $$ \lim_{p\to0}\left(\frac{2^{p+1}-1}{p+1}\right)^{\!1/p} $$ and it's better to compute the limit of the log: $$ \lim_{p\to0}\frac{\log(2^{p+1}-1)-\log(p+1)}{p} $$ You can do it separately: $$ \lim_{p\to0}\frac{\log(2^{p+1}-1)}{p} $$ is the derivative at $0$ of $f(p)=\log(2^{p+1}-1)$; since $$ f'(p)=\frac{1}{2^{p+1}-1}2^{p+1}\log2 $$ we have $f'(0)=2\log2$.

On the other hand, it should be well known that $$ \lim_{p\to0}\frac{\log(p+1)}{p}=1 $$ So your limit is $$ 2\log2-1=\log\frac{4}{e} $$ and your original limit is $4/e$.

Answer 2

$$X = \lim_{p\to 0} (2^p-1)^{1/p}$$

$$\log X = \lim_{p\to 0} 1/p \log(2^p-1) = -\infty$$ $$X=0$$

Answer 3

$$\lim_{p\to0}\int^{1}_{0} y^p dy$$ $$\lim_{p\to0} (y^{p+1}/p+1|_{1}^{2})^{1/p}$$ it reduces to $$\lim_{p\to0}[2^{p+1}/{p+1}-1/{p+1}]^{1/p}$$ now by applying limit $$\log X=\lim_{p\to0}1/p \log [2^{p+1}/{p+1}-1/{p+1}]$$ apply L hospitals

$$\frac{\frac{\log (2)(2^{p+1})(p+1)-(2^{p+1}-1)}{(p+1)^2}}{[2^{p+1}/{p+1}-1/{p+1}]}$$

now putting the $\lim p \to 0$

$$\ln X =\ln(4/e)$$

$$X=4/e$$

Answer 4

We have that

$$\int_0^1(1+x)^pdx=\begin{cases}\left.\log(1+x)\right|_0^1=\log2\;,&p=-1\\{}\\\left.\frac{(1+x)^{p+1}}{p+1}\right|_0^1=\frac1{p+1}\left(2^{p+1}-1\right)\;:&p\neq-1\end{cases}$$

so as $\;p\to 0\;$ we can safely assume the second option above, and then

$$\left(\int_0^1(1+x)^pdx\right)^{1/p}=\left(\frac{2^{p+1}-1}{p+1}\right)^{1/p}=$$

$$=\left[\frac1{(1+p)^{1/p}}\right](2^{p+1}-1)^{1/p}\xrightarrow[p\to0]{}\frac1e\cdot4=\frac4e$$

since

$$\lim_{p\to0}\log\left(2^{p+1}-1\right)^{1/p}=\lim_{p\to0}\frac{\log(2^{p+1}-1)}p\stackrel{\text{l'Hospital}}=$$

$$=\lim_{p\to0}\frac{2^{p+1}\log2}{2^{p+1}-1}=\frac{2\log2}1\implies\lim_{p\to0}(2^{p+1}+1)=e^{2\log2}=4$$

Answer 5

Apply $\ln $ to the expression to get

$$\frac{\ln [\int_0^1 (1+x)^p\, dx]}{p}.$$

Now as $p\to 0,$ the integral inside the brackets $\to 1,$ simply because $(1+x)^p \to 1$ uniformly. So we have a $0/0$ sitution and can apply L'Hopital. The quotient of derivatives is

$$\tag 1 \frac{\int_0^1 \ln (1+x)(1+x)^p\, dx}{\int_0^1 (1+x)^p\, dx}.$$

Here I've used "differentiation under the integral sign", which is simple to verify here. Again use $(1+x)^p \to 1$ uniformly to see the limit of $(1)$ is $\int_0^1 \ln (1+x), dx.$ That integral equals $\ln (4/e).$ Exponentiating back gives $4/e$ for the original limit.

Answer 6

In the same spirit as egreg's answer, let us consider $$A_p=\left(\frac{2^{p+1}-1}{p+1}\right)^{\!1/p}$$ Take logarithms $$\log(A_p)=\frac 1p \log\left(\frac{2^{p+1}-1}{p+1}\right)$$ Now, from definition, build Taylor series around $p=0$ to get $$\log\left(\frac{2^{p+1}-1}{p+1}\right)=p (2 \log (2)-1)+p^2 \left(\frac{1}{2}-\log ^2(2)\right)+O\left(p^3\right)$$ So $$\log(A_p)=(2 \log (2)-1)+p \left(\frac{1}{2}-\log ^2(2)\right)+O\left(p^2\right)$$ Now, using $A_p=e^{\log(A_p)}$ and Taylor series again $$A_p=\frac{4}{e}+\frac{2 -4\log ^2(2)}{e}p+O\left(p^2\right)$$ which shows the limit and also how it is approached.

For illustration purposes, let us try using $p=\frac 1{10}$. The exact result is $$A_{\frac 1 {10}}=\Big(\frac{10}{11} \left(2 \sqrt[10]{2}-1\right)\Big)^{10}\approx 1.474392296$$ while the approximation would give $$\frac{21-2 \log ^2(2)}{5 e}\approx 1.474394138$$ Using one more term in Taylor expansions would lead to an an approximate value $\approx 1.474392407$.