# Show the limit of parametric integral

Let's consider the parametric integral:

$$F:\mathbb{R}\to\mathbb{R}$$, where $$F(x):=\int\limits_0^1 \frac{1}{x}\left(e^{-x^2(1+t)}(1+t)^{-1}-(1+t)^{-1}\right)dt$$.

What is the value of the limit: $$\lim\limits_{x\to 0}F(x):=\lim\limits_{x\to 0}\int\limits_0^1 \frac{1}{x}\left(e^{-x^2(1+t)}(1+t)^{-1}-(1+t)^{-1}\right)dt?$$

Splitting up the integral into two summands and applying integration by parts at the first one doesn't get me anyhwere?! Any thoughts?

• Expand the exponential into a Taylor series. Most terms will vanish after taking the limit.
– Gary
Commented Feb 7, 2021 at 14:32
• @Gary, I am not sure if I have understood you hint correctly. You suggest to pull the limit into the integral and evaluate it, right? However, we don't know if this is possible. So I think we first must calculate the integral and then take the limit. Commented Feb 7, 2021 at 14:51

$$\lim_{x\to 0}\frac{\int_0^1 \left(\frac{e^{-x^2(t+1)}}{t+1}-\frac{1}{t+1}\right)\, dt}{x}=$$ $$\int_0^1\frac{dt}{t+1}=\log 2$$ Numerator becomes $$\int_0^1 \frac{e^{-x^2(t+1)}}{t+1}\, dt-\log 2$$ by Leibniz integral rule $$\frac{d}{dx}\int_0^1 \frac{e^{-x^2(t+1)}}{t+1}\, dt=\int_0^1 \frac{-2 (t+1) x e^{(-t-1) x^2}}{t+1} \, dt=\frac{2 e^{-2 x^2}}{x}-\frac{2 e^{-x^2}}{x}$$

Apply L'Hopital rule again:

$$=\lim_{x\to 0}\left(\frac{2 e^{-2 x^2}}{x}-\frac{2 e^{-x^2}}{x}\right)=\lim_{x\to 0}\left(4 e^{-x^2} x-8 e^{-2 x^2} x\right)=0$$

• To apply L'Hopital's rule me must show that the parametric integral in the nominator of your expression is differentiable (We don't know this). Even if we knew that the parametric integral in the nominator is differentiable what would be its derivative? Commented Feb 7, 2021 at 15:23
• @Philipp I have added a few lines. Hope it is more clear now. Commented Feb 7, 2021 at 15:35
• I don't see how the expression $$\lim\limits_{x\to 0}\frac{\int\limits_0^1 \frac{e^{-x^2(t+1)}}{t+1} dt-\log 2}{x}$$ yields an indeterminate form (e.g. $\frac{\infty}{\infty}$, or $\frac{0}{0}$) so that we can apply L'Hopital's rule. Can you elaborate on the integral $\int\limits_0^1 \frac{e^{-x^2(t+1)}}{t+1} dt$? Commented Feb 7, 2021 at 18:40
• @Philipp at $x=0$ it is $\int_0^1\frac{dt}{t+1}=\log 2$, that's why it is a $\frac{0}{0}$ form Commented Feb 7, 2021 at 18:50
• Oh my bad, it was pretty obvious :D thank you! Commented Feb 7, 2021 at 18:52

If you expand the exponential into its Taylor series and simplify the result, the integrand becomes $$\frac{1}{x}\left( {e^{ - x^2 (1 + t)} (1 + t)^{ - 1} - (1 + t)^{ - 1} } \right) = \sum\limits_{n = 0}^\infty {( - 1)^{n + 1} \frac{{x^{2n + 1} }}{{(n + 1)!}}(1 + t)^n } .$$ Integrating term-by-term (the series converges absolutely and uniformly in $$t\in [0,1]$$) yields $$F(x) = \sum\limits_{n = 0}^\infty {( - 1)^{n + 1} \frac{{x^{2n + 1} }}{{(n + 1)!}}\frac{{2^{n + 1} - 1}}{{n + 1}}} .$$ This clearly shows that $$\lim _{x \to 0} F(x) = 0$$.

• Have you used the following argument to justify your last step? As $\sum\limits_{n=0}^m{(-1)^{n+1}\frac{{x^{2n+1}}}{{(n+1)!}}\frac{{2^{n+1}-1}}{{n+1}}}$ is continuous the limit $\sum\limits_{n=0}^\infty{(-1)^{n+1}\frac{{x^{2n+1}}}{{(n+1)!}}\frac{{2^{n+1}-1}}{{n+1}}}$ inherits the continuity due to uniform convergence (We only consider those $x$ with $|x-0|\leq\epsilon$). Commented Feb 12, 2021 at 0:14
• Hence,$$\lim\limits_{x\to0}\sum\limits_{n=0}^\infty{(-1)^{n+1}\frac{{x^{2n+1}}}{{(n+1)!}}\frac{{2^{n+1}-1}}{{n+1}}}=\sum\limits_{n=0}^\infty{(-1)^{n+1}\frac{{(\lim\limits_{x\to0}x)^{2n+1}}}{{(n+1)!}}\frac{{2^{n+1}-1}}{{n+1}}}=0.$$(In fact, we have also used the continuity of $x^{2n+1}$) Correct? Commented Feb 12, 2021 at 0:14
• Yes, that is a reasoning behind it. Alternatively write $$\mathop {\lim }\limits_{x \to 0} F(x) = \mathop {\lim }\limits_{x \to 0} x\mathop {\lim }\limits_{x \to 0} \sum\limits_{n = 0}^\infty {( - 1)^{n + 1} \frac{{x^{2n} }}{{(n + 1)}}\frac{{2^{n + 1} - 1}}{{n + 1}}}$$ and note that the sum is uniformly bounded near the origin.
– Gary
Commented Feb 12, 2021 at 6:46