Inequality with definite integral I met some problems in getting the following limit
$$
\lim_{U\rightarrow\infty}\frac{U+1}{2^{U+1}-1}\int_0^1
y^{N-1}(y+1)^Udy
$$
for any fixed $N$ being natural numbers greater than or equal to 2.
Can someone also show $\frac{U+1}{2^{U+1}-1}\int_0^1
y^{N-1}(y+1)^Udy$ decreases or increases in $U$ for a fixed $N$?
At last, can someone show the following inequality?
$$
\left\{2\frac{N-1}{(U+N-1)N}\left[1-\frac{U+N}{2^{U+N}-1}\right](2^N-1)+1\right\}\frac{U+1}{2^{U+1}-1}\int_{0}^{1}y^{N-1}(y+1)^Udy>1
$$
where $U$ and $N$ are all natural numbers greater than or equal to 2. 
Thanks,   
 A: For large $U$, the integral can be estimated as
\begin{eqnarray*}
\int_{0}^{1}y^{N - 1}\left(1 + y\right)^{U}\,{\rm d}y
& = &
2^{U}\int_{0}^{1}\left(1 - \epsilon\right)^{N - 1}\left(1 - {\epsilon \over 2}\right)^{U}
{\rm d}\epsilon
=
2^{U}\int_{0}^{1}{\rm e}^{\left(N - 1\right)\ln\left(1 - \epsilon\right)\
                          +\ U\ln\left(1 - \epsilon/2\right)}{\rm d}\epsilon
\\
& \approx &
2^{U}\int_{0}^{1}{\rm e}^{-\left(N - 1\ +\ U/2\right)\epsilon}\,{\rm d}\epsilon
=
2^{U}\ \
{{\rm e}^{-\left(N - 1\ +\ U/2\right)}  - 1 \over -\left(N - 1 + U/2\right)}
\approx
{2^{U + 1} \over U}
\end{eqnarray*}
Next term $\sim {2^{U} \over U^{2}}$
$$
\lim_{U \to \infty}{U + 1 \over 2^{U + 1} - 1}
\int_{0}^{1}y^{N - 1}\left(1 + y\right)^{U}\,{\rm d}y
=
\lim_{U \to \infty}\left({U + 1 \over 2^{U + 1} - 1}\,\times\,{2^{U + 1} \over U}\right)
=
1
$$
A: We can also use Watson's lemma to read off the asymptotics.  If we make the substitution
$$
y = 2e^{-t}-1
$$
then the integral takes the form
$$
\int_0^1 y^{N-1} (1+y)^U\,dy = 2^{U+1} \int_0^{\log 2} (2e^{-t}-1)^{N-1} e^{-(U+1)t}\,dt.
$$
By Watson's lemma this new integral has an asymptotic expansion in powers of $1/(U+1)$,
$$
\int_0^{\log 2} (2e^{-t}-1)^{N-1} e^{-(U+1)t}\,dt \approx \frac{1}{U+1} + \frac{2-2N}{(U+1)^2} + \cdots
$$
as $U \to \infty$, and from this we get the equivalence
$$
\int_0^1 y^{N-1} (1+y)^U\,dy \sim \frac{2^{U+1}}{U+1}
$$
as $U \to \infty$.  If you'd like you can use the second term of the asymptotic expansion to get the better estimate
$$
\frac{U+1}{2^{U+1}-1}\int_0^1 y^{N-1} (1+y)^U\,dy = 1 + \frac{2-2N}{U+1} + O\left(\frac{1}{U+1}\right)^2
$$
as $U \to \infty$.  The left-hand side is eventually monotonic, so from this we can gather that it is eventually decreasing if $N<1$ and eventually increasing if $N > 1$.
A: Thank you, Antonio and Felix. What you suggested is very helpful in proving the limit but these transformations seem to hold only for large $U$. I am wondering with finite natural numbers $U$ and $N$, whether there is a way to write $\int_0^1 y^{N-1}(y+1)^Udy$ in a tractable form so that we could judge whether $\left\{2\frac{N-1}{(U+N-1)N}\left[1-\frac{U+N}{2^{U+N}-1}\right](2^N-1)+1\right\}\frac{U+1}{2^{U+1}-1}\int_{0}^{1}y^{N-1}(y+1)^Udy>1$ or not.
Thanks,
Will
