How to prove that $\displaystyle I=\int_0^{2\pi}e^{-x^2}\cos(x)\mathrm{d}x>0$

Clearly, $\displaystyle I=\int_0^\pi \left(e^{-x^2}-e^{-(x+\pi)^2}\right)\cos(x)\mathrm{d}x$, but this does not help.

Or else, $$I=\int_0^{2\pi} e^{-x^2}\mathrm{d}\sin(x) =2\int_0^{2\pi} xe^{-x^2}\sin(x)\mathrm{d}x =2\int_0^\pi \left(xe^{-x^2}-(x+\pi)e^{-(x+\pi)^2}\right)\sin(x)\mathrm{d}x.$$ This time, $\sin(x)$ has a good sign, but $xe^{-x^2}$ has no monotonicity property.

Any ideas?

  • $\begingroup$ Another MSE 'clearly' that clearly needs elucidation. $\endgroup$ Commented Jul 21, 2023 at 4:24
  • $\begingroup$ @ParamanandSingh Thanks for catching that. Way off $\endgroup$
    – copper.hat
    Commented Jul 21, 2023 at 4:48
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    $\begingroup$ The key here is split the interval into two parts $[0,a],[a,2\pi]$ such that $|e^{-x^2}\cos x|$ is small in $[a, 2\pi]$. $\endgroup$
    – Paramanand Singh
    Commented Jul 21, 2023 at 4:50
  • $\begingroup$ I think the first quadrant of the unit circle dominates. $\endgroup$
    – Bob Dobbs
    Commented Jul 21, 2023 at 5:36
  • $\begingroup$ @RyszardSzwarc: $f(x)=e^{-x^2}-e^{-(x+\pi)^2}$ is increasing for very small $x$. $\endgroup$
    – Martin R
    Commented Jul 21, 2023 at 13:29

4 Answers 4


Use the inequality $\cos x\ge 1-\frac{x^2}2$ to write: $$ \int_0^{2\pi}e^{-x^2}\cos x\,{\rm d}x\ge\int_0^{2\pi}\left( e^{-x^2}-\frac{x^2}2 e^{-x^2}\right)\,{\rm d}x=A - B.$$ To show $A>B$, compute a lower bound for $A$ using the inequality $e^{-t}\ge 1-t$: $$A:=\int_0^{2\pi}e^{-x^2}\,{\rm d}x\ge\int_0^1 e^{-x^2}\,{\rm d}x\ge\int_0^1\left(1- x^2\right)\,{\rm d}x=\frac23.$$ Find an upper bound for $B$ using the fact $\int_0^\infty x^2e^{-x^2}\,{\rm d}x=\frac{\sqrt\pi}4$: $$B:=\frac12\int_0^{2\pi}x^2e^{-x^2}\,{\rm d}x\le\frac12\int_0^\infty x^2e^{-x^2}\,{\rm d}x=\frac{\sqrt\pi}8.$$

This approach also proves the non-trivial case ($t>\frac\pi2$) for the general result $$ \int_0^t e^{-x^2}\cos x\,{\rm d}x>0\qquad\text {for all $t>0$}. $$


The integral $I_1=\int_0^{\pi/2}f(x)\,dx$ is positive (as the integrand is positive in this interval) and $$I_1>\int_0^{\pi/4}f(x)\,dx>\frac{\pi}{4}\cdot \frac{e^{-\pi^2/16}}{\sqrt{2}}\tag{1}$$ as the integrand is decreasing in this interval and hence greater than $f(\pi/4)$.

Further $I_2=\int_{\pi/2}^{3\pi/2}f(x)\,dx$ is negative with $$|I_2|\leq \int_{\pi/2}^{3\pi/2}|f(x)|\,dx<\pi \cdot e^{-\pi^2/4}\tag{2}$$ and our job is done if we can show that RHS of $(1)$ is greater than that of $(2)$.

This requires us to prove that $e^{3\pi^2/16}>4\sqrt {2}$ or $$\frac{3\pi^2}{8}>5\log 2$$ This is possible by noting that $\log 2<0.7$ and $\pi>3.14$.

  • $\begingroup$ How do you get Inequalities (1) and (2)? $\endgroup$
    – Hans
    Commented Jul 21, 2023 at 15:18
  • $\begingroup$ @Hans: updated my answer with some details. The inequalities are not tricky but rather based on well known properties of integrals. $\endgroup$
    – Paramanand Singh
    Commented Jul 21, 2023 at 18:29
  • $\begingroup$ Yeah, I should have noticed earlier... $\endgroup$
    – Hans
    Commented Jul 22, 2023 at 2:18

Bounds for the integral on subintervals: $[0,\pi/4]: I_{11}>\frac1{\sqrt2}e^{-\pi^2/16}\\ [\pi/4,\pi/2]: I_{12}>(1-\frac1{\sqrt2})e^{-\pi^2/4}\\ [\pi/2,\pi]: I_{2}>-e^{-\pi^2/4}\\ [\pi,3\pi/2]: I_{3}>-e^{-\pi^2}\\ [3\pi/2,2\pi]: I_{4}>e^{-4\pi^2}\\ $

Adding these we have $$I=\int_0^{2\pi}e^{-x^2}\cos x\, dx>\frac1{\sqrt2}(e^{-\pi^2/16}-e^{-\pi^2/4})-e^{-\pi^2}> \frac1{\sqrt2}(e^{-0.64}-e^{-9/4})-e^{-9}>0. $$


The integral is the area under the curve.

The product of $e^{-x^2}$ and $\cos(x)$ is mostly positive in the interval given for x . Think of it pointwise, at every x in the interval. You may split the integral into the negative and positive area parts of $\cos(x)$ since the $e^{-x^2}$ Gaussian function is positive and compare or bound those pieces for a formal proof.

enter image description here

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    $\begingroup$ This is persuasion, not a proof. $\endgroup$ Commented Jul 21, 2023 at 4:27
  • $\begingroup$ answers are open to expert proof writers $\endgroup$
    – vallev
    Commented Jul 21, 2023 at 7:07
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    $\begingroup$ I would not consider $e^{x^2} \cos x$ as "mostly positive" from $0$ to $2\pi$. The product is negative on $(\pi/2, 3\pi/2)$ and positive on the other two open subintervals. $\endgroup$ Commented Jul 21, 2023 at 9:26
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    $\begingroup$ Nice graph which shows that the subintegral on the subinterval $[0,\frac\pi 4]$ dominates. And it is positive. $\endgroup$
    – Bob Dobbs
    Commented Jul 22, 2023 at 4:05

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