What is the value of this improper integral? $\lim_{x\rightarrow 0 } \dfrac{1}{x} \int_{x}^{2x} e^{-t^2}\,\mathrm dt$

Question

$$\lim_{x\rightarrow 0 } \dfrac{1}{x} \int_{x}^{2x} e^{-t^2}\,\mathrm dt$$

I don't have any idea to solve this integral.

Answer 1

Perhaps, an easy way is using l'Hopital's rule:

$$\lim_{x\to 0}\frac{\int_x^{2x}e^{-t^2}\,dt}{x}=\lim_{x\to 0} (2e^{-4x^2}-e^{-x^2})=1$$

How do I differentiate the numerator?

You use the rule

$$g(x)=\int_{a(x)}^{b(x)}f(t)\,dt\implies\\\implies g'(x)=b'(x)f(b(x))-a'(x)f(a(x))$$

which is valid whenever $f$ is continuous on a suitably big interval (say, $\mathbb R$) and $a,b$ are differentiable.

Indeed, $\int_{a(x)}^{b(x)}f(t)\,dt=F(b(x))-F(a(x))$ for a differentiable function such that $F'(x)=f(x)$. And you get the formula above using the chain rule.

Answer 2

If you don't want to use L'Hôpital's rule, you can rewrite the limit as $$ \lim_{h\rightarrow 0 } \frac{\int_{h}^{2h} e^{-t^2} dt-\int_{0}^{2\cdot0} e^{-t^2} dt}{h} $$ and recognize the derivative of $$ f(x):=\int_{x}^{2x} e^{-t^2} dt $$ at $x=0$, which is (by the chain rule and the Fundamental Theorem of Calculus) $$ f'(0)=2e^{0}-e^0=1 $$

Answer 3

Hint: Write down the series $1-t^2+\cdots$ for $e^{-t^2}$, integrate term by term, and divide by $x$.

Remark: If your question was about finding an antiderivative of $e^{-t^2}$ as a first step in solving the problem, that will not work. For $e^{-t^2}$ does not have an antiderivative that can be expressed in terms of elementary functions.

Answer 4

This is not an improper integral ($x\to 0$)

Now denote $f(x)=\int_0^xe^{-t^2}dt$

Keeping in mind $f(0)=0$ and $f'(x)=e^{-x^2}$, we're looking for:

$$\begin{align}\lim_{x\to 0}{f(2x)-f(x)\over x}&=2f'(0)-f'(0)\\&=1\end{align}$$

Answer 5

Solution based in squeezing: for $x>0$, $$e^{-4x^2} = \frac1x (2x-x)e^{-(2x)^2}\le\dfrac{1}{x}\int_{x}^{2x}e^{-t^2}\, dt\le\frac1x(2x-x) e^{-x^2} = e^{-x^2}.$$ The case $x<0$ is similar.

Answer 6

In a neighbourhood of the origin: $$ e^{-t^2} = 1-t^2 + o(t^3) \tag{1}$$ hence: $$ \int_{x}^{2x}e^{-t^2}\,dt = x-\frac{7}{3}x^3+o(x^4) \tag{2}$$ and: $$\frac{1}{x}\int_{x}^{2x}e^{-t^2}\,dt = \color{red}{1}-\frac{7}{3}x^2+o(x^3).\tag{3}$$

Answer 7

More generally, for small $x$, if $f$ is differentiable at zero, $f(x) \approx f(0)+xf'(0) $ so

$\begin{array}\\ \int_{ax}^{bx} f(t) dt &\approx \int_{ax}^{bx} (f(0)+tf'(0)) dt\\ &= (tf(0) + t^2f'(0)/2)\mid_{ax}^{bx}\\ &= (x(b-a)f(0) + x^2(b^2-a^2)f'(0)/2)\\ \text{or}\\ \frac1{x}\int_{ax}^{bx} f(t) dt &\approx (b-a)f(0) + x(b^2-a^2)f'(0)/2)\\ &\to (b-a)f(0) \qquad\text{as } x \to 0\\ \end{array} $

If $f(x) = e^{-x^2}$, $a=1, b=2$, then $f(0) = 1$ so the limit is $(2-1)1 = 1 $.