Limit of characteristic function I have a characteristic function defined as the following:
$$\phi(\frac{t}{N})^N = (\alpha E_{\alpha+1}(-\frac{iLt}{N}))^N$$
where $E_n(z)$ is the $E_n$ function having the form $E_n(z) = \int^\infty_1 \frac{e^{-zm}dm}{m^n}$. And the limit is supposed to equal:
$$ \lim_{N \to \infty} (\alpha \int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha+1}})^N= e^{\frac{\alpha}{\alpha-1}iLt}.$$
for $\alpha > 1$ and $N$ $\in$ $\mathbb{N}$ and $L$ is a positive constant. I don't see why this limit holds, any ideas?
 A: Whenever we say "bounded", we mean it is bounded by a constant that doesn't depend on $N$.
Integrate by parts twice:
$$ \alpha \int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha+1}} $$
$$ = - \left[ \frac{e^{\frac{-iLtm}{N}}}{m^{\alpha}} \right]^\infty_1  - \frac{iLt}{N}\int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha}} $$
$$ = e^{\frac{-iLtm}{N}} + \left[ \frac{iLt}{N}\frac1{\alpha-1}\frac{e^{\frac{-iLtm}{N}}}{m^{\alpha-1}} \right]^\infty_1 + \left(\frac{iLt}{N}\right)^2\frac1{\alpha-1}\int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha-1}} $$
$$ = e^{\frac{-iLtm}{N}} - \frac{iLt}{N}\frac1{\alpha-1}e^{\frac{-iLtm}{N}} + \left(\frac{iLt}{N}\right)^2\frac1{\alpha-1}\int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha-1}} $$
If $\alpha>2$, then $\int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha-1}}$ is bounded, so what if $1 < \alpha \le 2$?
$$ \int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha-1}} = N^{2-\alpha}\int^\infty_{1/N} \frac{e^{-iLtm}dm}{m^{\alpha-1}} $$
$$ = N^{2-\alpha}\int^1_{1/N} \frac{e^{-iLtm}dm}{m^{\alpha-1}}
+ N^{2-\alpha}\int^\infty_{1} \frac{e^{-iLtm}dm}{m^{\alpha-1}} .$$
If $\alpha < 2$, then $\int^1_{1/N} \frac{e^{-iLtm}dm}{m^{\alpha-1}}$ is bounded by $\int^1_{0} \frac{1}{m^{\alpha-1}}$, which is bounded by the $p$-test.  If $\alpha = 2$, a similar argument shows it is bounded by $\log N$.
As for the other part, we integrate by parts the other way:
$$ \int^\infty_{1} \frac{e^{-iLtm}dm}{m^{\alpha-1}} $$
$$ =\left[ \frac1{-iLtm}\frac{e^{-iLtm}}{m^{\alpha-1}} \right]^\infty_{1} - \frac{\alpha-1}{-iLtm}\int^\infty_{1} \frac{e^{-iLtm}dm}{m^{\alpha}} $$
and again we get something bounded.
Putting it all together, we get
$$ \alpha \int^\infty_1 \frac{e^{\frac{-iLtm}{N}}dm}{m^{\alpha+1}} = e^{\frac{-iLtm}{N}}\left(1 - \frac{iLt}{N}\frac1{\alpha-1} + o(N^{-1})\right) .$$
