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?

  • $\begingroup$ Mathematica gives the analytic answer that first integral is larger. $\endgroup$ Jul 14, 2023 at 17:21
  • 1
    $\begingroup$ I don't see any apparent flaw in your proof. $\endgroup$
    – Bruno B
    Jul 14, 2023 at 18:08

2 Answers 2


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 $f(0) = g(0).$


Here is an elementary solution.

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .