I got two problems asking for the proof of the limit:
Prove the following limit:
$$\sup_{x\ge 0}\ x e^{x^2}\int_x^\infty e^{-t^2} \, dt={1\over 2}.$$
and,
Prove the following limit:
$$\sup_{x\gt 0}\ x\int_0^\infty {e^{-px}\over {p+1}} \, dp=1.$$
I may feel that these two problems are of the same kind. World anyone please help me with one of them and I may figure out the other one? Many thanks!
Let
$$ f(x)=\ x e^{x^2}\int_x^\infty e^{-t^2} \implies f(x)=\ x e^{x^2}g(x).$$
We can see that $ f(0)=0 $ and $f(x)>0,\,\, \forall x>0$. Taking the limit as $x$ goes to infinity and using L'hobital's rule and Leibniz integral rule yields
$$ \lim_{ x\to \infty } xe^{x^2}g(x) = \lim _{x\to \infty} \frac{g(x)}{\frac{1}{xe^{x^2}}}=\lim_{x \to \infty} \frac{g'(x)}{\frac{1}{(xe^{x^2})'}}=\lim_{x \to \infty} \frac{-e^{-x^2}}{{-{\frac {{{\rm e}^{-{x}^{2}}} \left( 2\,{x}^{2}+1 \right) }{{x}^{2}}}}} =\frac{1}{2}. $$
After a suitable substitution, both limits are easily handled by Dominated Convergence or Monotone Convergence.
The First Integral
Substituting $t\mapsto\sqrt{t+x^2}$, $$ \begin{align} xe^{x^2}\int_x^\infty e^{-t^2}\,\mathrm{d}t &=xe^{x^2}\int_0^\infty e^{-t-x^2}\frac{\mathrm{d}t}{2\sqrt{t+x^2}}\\ &=\frac12\int_0^\infty e^{-t}\frac{\mathrm{d}t}{\sqrt{1+t/x^2}}\tag{1} \end{align} $$ Equation $(1)$ shows that the integrand is increasing with $x$ and by Dominated or Monotone Convergence, the limit is $$ \frac12\int_0^\infty e^{-t}\,\mathrm{d}t=\frac12\tag{2} $$
The Second Integral
Substituting $p\mapsto p/x$, $$ \begin{align} x\int_0^\infty\frac{e^{-px}}{p+1}\,\mathrm{d}p &=\int_0^\infty\frac{e^{-p}}{1+p/x}\,\mathrm{d}p\tag{3} \end{align} $$ Equation $(3)$ shows that the integrand is increasing with $x$ and by Dominated or Monotone Convergence, the limit is $$ \int_0^\infty e^{-p}\,\mathrm{d}p=1\tag{4} $$
For the first problem, use integration by parts, followed by crude estimation.
Let $u=\frac{1}{t}$ and $dv=te^{-t^2}\,dt$. Then $du=-\frac{1}{t^2}\,dt$ and we can take $v=-\frac{1}{2}e^{-t^2}$. Thus our integral is equal to $$\frac{1}{x}\cdot \frac{1}{2}e^{-x^2}-\int_x^\infty \frac{1}{2t^2}e^{-t^2}\,dt.$$ Multiplying $\frac{1}{x}\cdot\frac{1}{2}e^{-x^2}$ by $xe^{x^2}$ gives us the main term. It remains to check that $\int_x^\infty \frac{1}{2t^2}e^{-t^2}\,dt$ is small enough that multiplying by $xe^{x^2}$ gives a small result.
We have $$\int_x^\infty \frac{1}{2t^2}e^{-t^2}\,dt \lt \frac{1}{2x^2}\int_x^\infty e^{-t^2}\,dt.$$ We can bound $\int_x^\infty e^{-t^2}\,dt$ by integrating from $x$ to $x+1$, and from $x+1$ to $\infty$. The integral from $x$ to $x+1$ is $\lt e^{-x^2}$.
For $x+1$ to $\infty$, make the change of variable $t=s+1$. We want $\int_{s=x}^\infty e^{-s^2-2s-1}\,ds$, which is less than $e^{-x^2}\int_x^\infty e^{-2s-1}\,ds$. The remaining integral is bounded by a constant.
I'll do the second one. And it appears that André Nicolas did the first one while I was writing, so everything is fine.
By integration by parts, $$ x\int_0^{+\infty}\frac{e^{-px}}{p+1}dp=1-\int_0^{+\infty}\frac{e^{-px}}{(p+1)^2}dp\leq 1 $$ for all $x>0$.
Now by Lebesgue dominated convergence theorem, the rhs tends to $1-0=1$ as $x\rightarrow +\infty$.
So the sup over $x>0$ is indeed $1$.
The main things I'd add to the other answers is to explicitly apply the definition of a limit, rather than leave it all to higher-level theorems, since the OP said the problem was proving the limits.
$$\lim_{x \rightarrow \infty} f(x) = L \Leftrightarrow \forall \; \varepsilon > 0 \; \exists \; N > 0 \ni x > N \Rightarrow \left| f(x) - L \right| < \varepsilon$$
$$\;\\ x e^{x^2}\int_x^\infty e^{-t^2} \, dt = \lim_{b \rightarrow \infty} x e^{x^2}\int_x^b e^{-t^2} \, dt = L \Leftrightarrow \\ \forall \; \varepsilon > 0 \; \exists \; N > 0 \ni b > N \Rightarrow \left| e^{x^2}\int_x^b e^{-t^2} \, dt - L \right| < \varepsilon$$
$$\;\\ \text{Let} L = {1\over2}$$
$$\text{Show:} \; \forall \; \varepsilon > 0 \; \exists \; N > 0 \ni b > N \Rightarrow \left| x e^{x^2}\int_x^b e^{-t^2} \, dt - {1\over2} \right| < \varepsilon$$
And the rest of the whole rigmarole of finding some expression for $N$ containing $\varepsilon$ and $x$ that you can prove is always big enough (and then doing so).
