# How to prove that $\int_0^{2\pi}e^{-x^2}\cos xdx>0$

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?

• Another MSE 'clearly' that clearly needs elucidation. Commented Jul 21, 2023 at 4:24
• @ParamanandSingh Thanks for catching that. Way off Commented Jul 21, 2023 at 4:48
• 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 $[a, 2\pi]$. Commented Jul 21, 2023 at 4:50
• I think the first quadrant of the unit circle dominates. Commented Jul 21, 2023 at 5:36
• @RyszardSzwarc: $f(x)=e^{-x^2}-e^{-(x+\pi)^2}$ is increasing for very small $x$. Commented Jul 21, 2023 at 13:29

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 $$\pi>3.14$$.

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

• This is persuasion, not a proof. Commented Jul 21, 2023 at 4:27
• answers are open to expert proof writers Commented Jul 21, 2023 at 7:07
• 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. Commented Jul 21, 2023 at 9:26
• Nice graph which shows that the subintegral on the subinterval $[0,\frac\pi 4]$ dominates. And it is positive. Commented Jul 22, 2023 at 4:05