Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to evaluate: $$ \int_0^\infty e^{-x^2} \cos^n(x) dx$$

Someone has posted this question on fb. I hope it's not duplicate.

share|improve this question
yeah I am looking for closed expression. The $\cos^n(x)$ be expanded in terms of $\cos (kx) $ and evaluated in terms of series. –  Santosh Linkha Feb 23 '13 at 9:44
Mh Mathematica doesn't give a closed expression, but for some tries he gave me an analytical result –  Dominic Michaelis Feb 23 '13 at 10:27
Nice question (+1) –  Chris's sis Feb 23 '13 at 10:30
what is fb ???? –  0x90 Feb 23 '13 at 10:35
@ox90 short for facebook –  Dominic Michaelis Feb 23 '13 at 10:36

2 Answers 2

up vote 7 down vote accepted

I found a way to do it for $n \in \mathbb{N}$. We begin with

$$\cos^n(x)=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n = \frac{1}{2^n e^{inx}}(1+e^{2ix})^n = \frac{1}{2^n e^{inx}}\sum_{r=0}^n \binom{n}{r}e^{2irx}$$


$$\begin{aligned}\int_{-\infty}^\infty e^{-x^2}\cos^n(x)dx &=\int_{-\infty}^\infty e^{-x^2}\frac{1}{2^n e^{inx}}\sum_{r=0}^n \binom{n}{r}e^{2irx} dx \\ &=\frac{1}{2^n}\sum_{r=0}^n \binom{n}{r}\int_{-\infty}^\infty e^{-x^2+(2ir-in)x}dx\end{aligned}$$

Here we can use the formula, $\int_{-\infty}^\infty e^{-x^2+bx+c}dx=\sqrt{\pi}e^{b^2/4+c}$. Applying it gives

$$\int_{-\infty}^\infty e^{-x^2}\cos^n(x)dx= \frac{\sqrt{\pi}}{2^n}\sum_{r=0}^n \binom{n}{r}\exp\left({\frac{-(2r-n)^2}{4}}\right)$$

The integrand is even, so

$$\int_0^\infty e^{-x^2}\cos^n(x)dx=\boxed{\displaystyle \frac{\sqrt{\pi}}{2^{n+1}}\sum_{r=0}^n \binom{n}{r}\exp\left({\frac{-(2r-n)^2}{4}}\right)}$$

share|improve this answer
all the functions are symmetric so divide it by two –  0x90 Feb 23 '13 at 10:34
@Dominic Michaelis: I corrected my answer. –  Integrals and Series Feb 23 '13 at 10:34
It should be $2^{n+1}$. –  Chris's sis Feb 23 '13 at 10:35
@Chris's sister and Pals:Thank you for pointing that out. I think I might have made a typo. –  Integrals and Series Feb 23 '13 at 10:38
@ShobhitBhatnagar Your final expression doesn't give correct results. For $n = 3$, Mathematica says the integral evaluates to $\frac{\sqrt{\pi }}{8 e^{9/4}}+\frac{3 \sqrt{\pi }}{8 \sqrt[4]{e}}$. However, yours gives $\frac{3 \sqrt{\pi }}{16 e^{81/4}}+\frac{\sqrt{\pi }}{16 e^{225/4}}+\frac{\sqrt{\pi }}{4 e^{9/4}}$. Probably a small mistake somewhere. –  Ayman Hourieh Feb 23 '13 at 10:42

Using the power reduction formula for $\cos$ for odd powers:

\begin{align} I &= \int_0^\infty e^{-x^2} \cos^n(x) \, dx \\ &= \int_0^\infty e^{-x^2} \left( \frac{2}{2^n} \sum_{k=0}^{(n-1)/2} \binom{n}{k} \cos{((n-2k)x)} \right) \, dx \\ &= \frac{2}{2^n} \sum_{k=0}^{(n-1)/2} \binom{n}{k} \int_0^\infty e^{-x^2} \cos{((n-2k)x)} \, dx \\ \end{align}

The inner integral is a generalization of the Gaussian integral and can be evaluated using differentiation under the integral sign: $$ \int_0^\infty e^{-x^2} \cos{((n-2k)x)} \, dx = \frac{1}{2} \sqrt{\pi} e^{-\frac{1}{4}(n-2 k)^2} $$

Therefore, we have:

$$ I = \frac{\sqrt{\pi}}{2^n} \sum_{k=0}^{(n-1)/2} \binom{n}{k} e^{-\frac{1}{4}(n-2 k)^2} \qquad n \text{ odd} $$

I don't think there is a nice closed form for this sum.

As for even powers, the same method yields:

\begin{align} I &= \int_0^\infty e^{-x^2} \cos^n(x) \, dx \\ &= \int_0^\infty e^{-x^2} \left( \frac{1}{2^n} \binom{n}{n/2} + \frac{2}{2^n} \sum_{k=0}^{n/2-1} \binom{n}{k} \cos{((n-2k)x)} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+1}} \binom{n}{n/2} + \frac{2}{2^n} \sum_{k=0}^{n/2-1} \binom{n}{k} \int_0^\infty e^{-x^2} \cos{((n-2k)x)} \, dx \end{align}


$$ I = \frac{\sqrt{\pi}}{2^{n+1}} \binom{n}{n/2} + \frac{\sqrt{\pi}}{2^n} \sum_{k=0}^{n/2-1} \binom{n}{k} e^{-\frac{1}{4}(n-2 k)^2} \qquad n \text{ even} $$

share|improve this answer
I think the top is $\lfloor {n \over 2} \rfloor$ let's see what pops up –  Santosh Linkha Feb 23 '13 at 11:12
@experimentX The expression in my answer is for odd powers. It's slightly different for even powers. I'll add it when I get a chance. –  Ayman Hourieh Feb 23 '13 at 12:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.