How to show that this integral is greater than the other integral?

Question

How can we show that $$\int_{x-b}^{x-a}e^{-t^2}dt > \int_{x+a}^{x+b}e^{-t^2}dt$$ for $x>0$, $0 < a < b$?

EDIT: I can show the case for when $x-b > 0$. First, $e^{-t^2}$ is monotonically decreasing on the range $t>0$ since the derivative $-2te^{-t^2} < 0$ on that range. Now, by the Extreme value theorem, $e^{-t^2}$ attains a minimum at $t=x-a$ in the range $[x-b, x-a$]. Hence, $$ \int_{x-b}^{x-a}e^{-t^2}dt > \int_{x-b}^{x-a}e^{-(x-a)^2}dt = (b-a)e^{-(x-a)^2} \tag{1} $$. For similar reasons, we can show that $$\int_{x+a}^{x+b}e^{-t^2}dt < \int_{x+a}^{x+b}e^{-(x+a)^2}dt = (b-a)e^{-(x+a)^2}\tag{2} $$ We can show that $(b-a)e^{-(x-a)^2} > (b-a)e^{-(x+a)^2}$ by demonstrating that $\frac{(b-a)e^{-(x-a)^2}}{(b-a)e^{-(x+a)^2}} > 1$. We have $$ \begin{split} \frac{(b-a)e^{-(x-a)^2}}{(b-a)e^{-(x+a)^2}} &= \frac{e^{-(x-a)^2}}{e^{-(x+a)^2}}\\ &=\frac{e^{2ax}}{e^{-2ax}}\\ &=e^{2ax-(-2ax)}\\ &=e^{4ax} > 1 \end{split} $$ for $x>0$ and $a>0$. Therefore, from (1) and (2) $\Rightarrow \int_{x-b}^{x-a}e^{-t^2}dt > \int_{x+a}^{x+b}e^{-t^2}dt$

Answer 1

Hint Here's an outline of a solution that dispenses with calculus as soon as possible.

1. Substitute $u = -t + x$ in the left-hand integral and $u = t - x$ in the right-hand integral, so that we are comparing two integrals over the same interval; we now want to show: $$\int_a^b e^{-(u - x)^2} du > \int_a^b e^{-(u + x)^2} du .$$
2. Observe that since $a < b$, to show the inequality it suffices to show that the corresponding inequality holds between the integrands, that is, that $$e^{-(u - x)^2} > e^{-(u + x)^2}$$ for all $u$ in the interval $[a, b]$ of integration.

3. One way to establish the inequality in (2) is to multiply both sides by the reciprocal $e^{(u + x)^2}$ of its right-hand side, which is positive for all $u$ and so does not change the direction of the inequality. Simplify.

Answer 2

I'm adding the rest of my solution here for completion.

First, we can show that $$ \int_{x}^{x+\alpha} f(t)dt > \int_{x+\gamma}^{x+\gamma+\alpha} f(t) dt $$ for $f(t)$ a monotonically decreasing function on $t \in [0; +\infty)$ such that $f(t) > 0$ on that range, $x \geq 0$ and $\alpha, \gamma > 0$.

Case I: $x+\gamma \geq x+\alpha$

By the Extreme value theorem, $f(t)$ attains a minimum at $x+\alpha$ in the range $[x; x+\alpha]$, hence: $$ \int_{x}^{x+\alpha} f(t)dt > \alpha f(x+\alpha) $$ Similarly, we can show that $$ \begin{split} \int_{x+\gamma}^{x+\gamma+\alpha}f(t)dt < \alpha f(x+\gamma) \end{split} $$ Since $f(t)$ is monotonically decreasing, $\alpha f(x+\gamma) \leq \alpha f(x+\alpha)$. Therefore, we have $$ \begin{align} \int_{x}^{x+\alpha} f(t)dt > \alpha f(x+\alpha) &\geq \alpha f(x+\gamma) > \int_{x+\gamma}^{x+\gamma+\alpha}f(t)dt \\ \int_{x}^{x+\alpha} f(t)dt &> \int_{x+\gamma}^{x+\gamma+\alpha}f(t)dt\tag{3} \end{align} $$

Case II: $x+ \gamma < x + \alpha$. Note that $x+ \gamma +\alpha > x + \alpha$

Here we have $$ \int_{x}^{x+\alpha}f(t)dt = \int_{x}^{x+\gamma} f(t)dt + \int_{x+\gamma}^{x+\alpha}f(t)dt $$ Let $\int_{x+\gamma}^{x+\alpha}f(t)dt = k$. We can then write the above integral as $$ \int_{x}^{x+\alpha}f(t)dt = k + \int_{x}^{x+\gamma} f(t)dt\tag{4} $$ Similarly, we can show that $$ \int_{x+\gamma}^{x+\gamma+\alpha}f(t)dt = k + \int_{x+\alpha}^{x+\gamma+\alpha}f(t)dt\tag{5} $$ Using $(3)$ we can show that $\int_{x}^{x+\gamma}f(t)dt > \int_{x+\alpha}^{x+\gamma+\alpha}f(t)dt$. Hence, $$ k+\int_{x}^{x+\gamma}f(t)dt > k+\int_{x+\alpha}^{x+\gamma+\alpha}f(t)dt\tag{6} $$ From $(4)$, $(5)$ and $(6) \Rightarrow$ $$ \int_{x}^{x+\alpha} f(t)dt > \int_{x+\gamma}^{x+\gamma+\alpha} f(t) dt\tag{7} $$

Now, going back to the original problem. In the case of $x - b < 0 \leq x - a$ we have $$ \begin{split} \int_{x-b}^{x-a}e^{-t^2}dt &= \int_{x-b}^{0}e^{-t^2}dt + \int_{0}^{x-a}e^{-t^2}dt\\ &=\int_{0}^{b-x}e^{-t^2}dt + \int_{0}^{x-a}e^{-t^2}dt \end{split} $$ and $$ \begin{split} \int_{x+a}^{x+b} e^{-t^2}dt &= \int_{x+a}^{2x}e^{-t^2}dt + \int_{2x}^{x+b} e^{-t^2}dt \end{split} $$ Using $(3)$ we can show that $$ \int_{0}^{x-a} e^{-t^2}dt > \int_{x+a}^{2x} e^{-t^2}dt\tag{8} $$ Similarly, using $(7)$ we can show that $$ \int_{0}^{b-x} e^{-t^2} dt > \int_{2x}^{x+b} e^{-t^2} dt\tag{9} $$ Combining $(8)$ and $(9) \Rightarrow$ $$ \int_{x-b}^{x-a}e^{-t^2}dt > \int_{x+a}^{x+b} e^{-t^2}dt $$

Lastly, we have the case of $x-b < x- a < 0$. Here, we have $$ \int_{x-b}^{x-a}e^{-t^2}dt = \int_{a-x}^{b-x}e^{-t^2}dt\tag{10} $$ because $e^{-t^2}$ is an even function. Directly applying $(7)$ here gives that $\int_{a-x}^{b-x}e^{-t^2}dt > \int_{x+a}^{x+b}e^{-t^2}dt$. Therefore, that, combined with $(10)$ gives $$ \int_{x-b}^{x-a}e^{-t^2}dt > \int_{x+a}^{x+b} e^{-t^2}dt $$