<s>[INST] Bounding a somewhat complicated integral (exponential of a polynomial) I am interested in bounding the following integral, where $a>0$ is a constant:
$$\int\limits_0^a \exp\left(\left(x^2 - \frac{2a^4+3}{4a^2}\right)^2\right) dx$$
I first conjectured that 
$$\int\limits_0^a \exp\left(\left(x^2 - \frac{ba^4+c}{a^2}\right)^2 \right) dx \le a^{-1}\exp \left( \left( \frac{ba^4+c}{a^2}\right)^2 \right)$$
when $c \ge 0$ and $b > 0$. However, this is false (for example, take $c = 0$, $b = 9/20$, and plug in $a = 2$). However, this inequality does appear to hold when $b \ge \frac{1}{2}$, from the numerical tests I have done. I am not sure how to prove it, however. The expressions that I have seen for exponentials of quartics are quite complicated and do not easily admit an upper bound (see for instance Exponential of a Quartic). Does anyone have any ideas?
[Note: In fact, what I'm really interested in is the regime $a \rightarrow \infty$, but so far the upper bound I have guessed appears to hold up for all $a > 0$.]
Updated wording: I am not just looking for any bound on this integral, but I indeed would like to prove the bound that I have proposed above. This was not made clear in my original post; I apologize for that.
 [/INST] Another approach I can think of that could provide more insight would be computing a Taylor Remainder for the expansion around $a=0$ of:
$$
F(a) = \int_0^a f(a,x)dx = \int_0^a \exp(g(a,x))dx 
$$
$$
F(a) = F(0) + F'(a) + F''(\xi)/2
$$
Naturally $F(0) = 0$. So:
For that, we can use Leibniz Integral Rule:
$$
\frac{dF}{da} |_{x=a} = f(a,a) \times 1 + \int_0^a \frac{\partial f(a,x)}{\partial a}dx  = W(a)
$$
Computing $f(a,a)$:
$$
f(a,a) = \exp \left(\frac{(1-b)a^4 -c}{a^2} \right)^2
$$
Now the partial derivative:
$$
\frac{\partial f(a,x)}{\partial a} = f(a,x) \times \frac{\partial g(a,x)}{\partial a} = f(a,x) \times 2 \left( \left( x^2 - \frac{ba^4+c}{a^2}\right) \left( -2ab +2\frac{c}{a^3}\right) \right) 
$$
Well... I'll just use Mathematica:
$$
W(a) = \int_0^a \frac{\left(a^2-3\right) e^{\left(\frac{3}{4 a^4}-\frac{1}{2 a^2}+x^2\right)^2}
   \left(4 a^4 x^2-2 a^2+3\right)}{2 a^9} \, dx+e^{\frac{\left(4 a^6-2 a^2+3\right)^2}{16
   a^8}}
$$
The Remainder term will look like this:
$$
\frac{e^{\left(\frac{3}{4 \xi ^4}+\xi ^2-\frac{1}{2 \xi ^2}\right)^2} \left(\xi ^6+\xi
   ^2-3\right) \left(4 \xi ^6-2 \xi ^2+3\right)}{\xi ^9}+\int_0^{\xi }
   \frac{e^{\left(\frac{3-2 \xi ^2}{4 \xi ^4}+x^2\right)^2} \left(-162 \xi ^2-24 \xi ^{14}
   x^2-36 \xi ^6 \left(8 x^2+1\right)+9 \xi ^4 \left(24 x^2+13\right)+4 \xi ^{12} \left(4
   x^4+30 x^2+5\right)-2 \xi ^{10} \left(48 x^4+8 x^2+63\right)+2 \xi ^8 \left(72 x^4+60
   x^2+83\right)+81\right)}{4 \xi ^{18}} \, dx
$$
So, I call it quits on this attempt, but I do admit it is starting to look plausible, but for a very limited range on $a$, such that $a^{-1}$ is actually a big number and thus $a^{-1} K $ for big enough $K$ can be an actual bound... But thinking like this...

Yeah... Nope, I don't think this would work...
</s>