Comparing two definite integrals analytically

How do you find which one is greater analytically: $$\displaystyle \int_{0}^{\int_0^1e^{-x^2}\mathrm dx} e^{x^2}\mathrm dx$$ or $$\displaystyle \int_{0}^{\int_0^1e^{x^2}\mathrm dx} e^{-x^2}\mathrm dx$$?

SMMC is an international undergrad level math competition (Eastern counterpart of the Putnam). This is a sample question from their site. There's a solution put up there which I'm presenting here in a more detailed manner.

Define as follows, $$f(t):=\displaystyle \int_{0}^{\int_0^t e^{-x^2}\mathrm dx} \exp(x^2)\mathrm dx\tag{01}$$ $$g(t):=\displaystyle \int_{0}^{\int_0^t e^{x^2}\mathrm dx} \exp(-x^2)\mathrm dx\tag{02}$$ Clearly, $$f(0)=g(0)=0$$. We intend to compare $$f(1)$$ and $$g(1)$$. Differentiating w.r.t. $$t$$ (use the fundamental theorem of calculus and the chain rule), $$f'(t)=\displaystyle \exp\left[\left(\int_0^t e^{-x^2}\mathrm dx\right)^2\right]\cdot e^{-t^2}\tag{03}$$ $$g'(t)=\displaystyle \exp\left[-\left(\int_0^t e^{x^2}\mathrm dx\right)^2\right]\cdot e^{t^2}\tag{04}$$ It looks like both $$f$$ and $$g$$ are increasing functions because $$f'$$ and $$g'$$ are $$+$$ve for all $$t$$. Given their initial value is same i.e., $$0$$ at $$t=0$$, it’s sufficient to check which one of them grows faster to compare their values at $$t=1$$. $$\displaystyle \frac{f'(t)}{g'(t)}=\exp\left[\left(\int_0^t e^{-x^2}\mathrm dx\right)^2+\left(\int_0^t e^{x^2}\mathrm dx\right)^2-2t^2\right]\tag{05}$$ By A.M.-G.M. inequality, we have: $$\displaystyle\left(\int_0^t e^{-x^2}\mathrm dx\right)^2+\left(\int_0^t e^{x^2}\mathrm dx\right)^2\\ \displaystyle \geq 2\int_0^t e^{-x^2}\mathrm dx\cdot \int_0^t e^{x^2}\mathrm dx\tag*{}$$ Notice that $$e^{x^2}$$ is increasing and $$e^{-x^2}$$ is decreasing over the interval $$(0, t)$$. We can apply the continuous analog of Chebyshev’s sum inequality.

If $$f(x)$$ is an increasing function and $$g(x)$$ is a decreasing function (or vice-versa) over the interval $$(a,b)$$, we have the following inequality: $$\displaystyle \frac{1}{b-a}\int_a^b f(x)\cdot g(x)\ \mathrm dx \\ \displaystyle \leq \left(\frac{1}{b-a}\int_a^b f(x)\mathrm dx\right)\cdot\left(\frac{1}{b-a}\int_a^b g(x)\mathrm dx\right)\tag*{}$$ The inequality is reversed if $$f(x)$$ and $$g(x)$$ are both increasing or both decreasing. This is valid for discrete sum as well, where instead of functions, we consider sequences.

$$\displaystyle \int_0^t e^{-x^2}\mathrm dx\cdot \int_0^t e^{x^2}\mathrm dx\\ \geq \displaystyle (t-0)\int_0^t e^{-x^2}\cdot e^{x^2}\mathrm dx =t^2 \tag*{}$$ Now we have established that: $$\displaystyle \left(\int_0^t e^{-x^2}\mathrm dx\right)^2+\left(\int_0^t e^{x^2}\mathrm dx\right)^2> 2t^2\tag{06}$$ The equality holds only if $$t=0$$. From $$(05)$$ and $$(06)$$, we have that: $$\displaystyle \frac{f'(t)}{g'(t)}>1 \text{ i.e., } f'(t)>g'(t)\tag*{}$$ $$\therefore$$ $$f$$ grows faster than $$g$$.

$$f(0) = g(0)$$ and hence, $$f(1)>g(1)$$. $$\blacksquare$$

I hope there is no mistake in my solution. Is there any alternate way to do this without making use of the sum inequality?

• Mathematica gives the analytic answer that first integral is larger. Jul 14, 2023 at 17:21
• I don't see any apparent flaw in your proof. Jul 14, 2023 at 18:08

The answer should follow by expanding the integrals in $$f'(t), g'(t)$$ by their taylor series. By elementary calculus, for any $$x \in \mathbb{R}$$, $$e^{x^2} = \sum_{n=0}^\infty \frac{x^{2n}}{n!} \qquad e^{-x^2} = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}.$$ Since the series are absolutely convergent, $$\int_0^t e^{x^2}dx = \sum_{n=0}^\infty \frac{t^{2n+1}}{n!(2n+1)} \qquad \int_0^t e^{-x^2}dx = \sum_{n=0}^\infty (-1)^n\frac{t^{2n+1}}{n!(2n+1)}.$$ From what you wrote above, \begin{align*}f'(t) = \exp\left( \left( \int_0^t e^{-x^2}dx\right)^2 - t^2\right) &= \exp\left( \sum_{n=0}^\infty \sum_{k=0}^n (-1)^k \frac{t^{2k+1}}{k!(2k+1)}(-1)^{n-k}\frac{t^{2(n-k)+1}}{(n-k)!(2(n-k)+1)} - t^2\right) \\ &= \exp\left( \sum_{n=0}^\infty \sum_{k=0}^n (-1)^n \frac{t^{2n+1}}{k!(2k+1)(n-k)!(2(n-k)+1)} - t^2\right). \end{align*} The work for $$g'(t)$$ is identical, and we end up with $$g'(t) = \exp\left( -\sum_{n=0}^\infty \sum_{k=0}^n \frac{t^{2n+1}}{k!(2k+1)(n-k)!(2(n-k)+1)} + t^2\right).$$ From this point, note that the $$n=0$$ term in the sums corresponds to the $$t^2$$ term. For $$t > 0$$, it is straightforward that $$f'(t) > g'(t)$$ because some of the terms in the exponential are nonnegative, while all the terms in $$g'(t)$$ are negative. This implies $$f(1) > g(1)$$ since $$f(0) = g(0).$$
Let $$p(t) = \int_{0}^{t}e^{-x^{2}}dx$$ and $$q(t)=\int_{0}^{t}e^{x^{2}}dx$$. Using the notation in the post, we wish to show $$f'(x) \ge g'(x)$$. It suffices to take logarithms (since they preserve inequalities), whence we wish to prove $$p\left(t\right)^{2}-t^{2} \ge -q\left(t\right)^{2}+t^{2}$$, i.e. that $$p\left(t\right)^{2}+q\left(t\right)^{2}-2t^{2} \ge 0$$. The LHS is zero at zero and is an even function, so it suffices to show the derivative is positive for $$x \ge 0$$, i.e. that $$2p(t)p'(t) + 2q(t)q'(t) - 4t \ge 0$$ for $$t \ge 0$$. We can bound each term below by their Taylor Series, so it suffices to prove $$2\left(1-t^{2}\right)\left(t-\frac{t^{3}}{3}\right)+2\left(1+t^{2}\right)\left(t+\frac{t^{3}}{3}\right)-4t \ge 0$$ for $$t \ge 0$$. However, the LHS is simply equal to $$\frac{4t^{5}}{3}$$ so we are done.