How to show that $\int_0^{|t|} s^{N-1} +s^N e^{s^{N^{\prime}}} ds\le\frac{|t|^N}{N}+|t|^{N+N^{\prime}}e^{|t|^{N^\prime}}$ Let $N\in\mathbb{N}, N\ge 2$ and let $N^{\prime}$ its Holder conjugate. I am trying to prove that the inequality
$$\int_0^{|t|} s^{N-1} +s^N e^{s^{N^{\prime}}} ds\le\frac{|t|^N}{N}+|t|^{N+N^{\prime}}e^{|t|^{N^\prime}}$$
holds true.
I proceeded in this way:
$$\int_0^{|t|} s^{N-1} +s^N e^{s^{N^{\prime}}} =\frac{|t|^N}{N} +\int_0^{|t|} s^N e^{s^{N^{\prime}}} ds.$$
Performing integration by parts, I have
$$\int_0^{|t|} s^N e^{s^{N^{\prime}}} ds=\frac{|t|^{N+1}}{N+1} e^{|t|^{N^{\prime}}}-\frac{N^\prime}{N+1}\int_0^{|t|}s^{N+N^\prime}e^{s^{N^{\prime}}} ds$$
but I don't know how to proceed from that.
Could someone please help?
Thank you in advance.
 A: Don't perform integration by parts with the terms you used : instead, note that $$
\frac{d}{dx}e^{x^{N'}} = e^{x^{N'}}N'x^{N'-1} 
$$
Using this one as an IBP part instead,
$$
\int_0^{|t|}s^Ne^{s^{N'}}ds = \frac{1}{N'}\int_0^{|t|} s^{N+1-N'}\left[e^{s^{N'}}N's^{N'-1}ds\right]
$$
Let $u= \frac{1}{N'}s^{N+1-N'}$ and $v = e^{s^{N'}}$, then this expression is equal to $\int_0^{|t|} u dv$, so
so we use IBP and get $$
\int_0^{|t|}s^Ne^{s^{N'}}ds = \left[\frac{1}{N'} e^{s^{N'}}s^{N+1-N'}\right]_0^{|t|} - \int_0^{|t|} e^{s^{N'}} [\frac{N+1-N'}{N'}s^{N-N'}ds] \\ \leq \frac{1}{N'}e^{|t|^{N'}}|t|^{N+1-N'} \leq e^{|t|^{N'}}|t|^{N+N'}
$$
because the integrand in the removed term is non-negative, and $N'\geq 1$, hence $1-N' \leq 0 \leq N'$ and $\frac{1}{N'} \leq 1$.
I think the inequality with $\frac{1}{N'}e^{|t|^{N'}}|t|^{N+1-N'}$ is probably tight. For example, take $N= N' = 2$,then some calculations done here show that the right asymptotic term is $xe^{x^2}$ up to a constant, which is equal to $N+1-N' = 1$ (and not close to $N + N' = 4$).
