# Integral of $\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)d\theta }$?

I am sorry if it does not fit here. I found some of the integral for the complementary error function e.g.

So far I did not find any integral regarding,

$$\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)d\theta }$$

Or, $$\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)\operatorname{erfc}\left( {\sin \left( {a + \theta } \right)} \right)d\theta }$$

1. Is it not possible to find the closed form of the integral of complimentary error function with trigonometric function inside.
2. Can anyone share me any integral of complimentary error function that has trigonometric function inside as argument?
• Firstly, you're integrating over $(0,2\pi)$, so $a$ does not matter. Secondly, $\text{erf}(x)$ is odd, so this implies $\int_0^{2\pi} \text{erf}(\cos x) dx = 0$. Thirdly, $\int_0^{2\pi} \text{erf}(\cos x) \text{erf}(\sin x) dx$ vanishes due to same reason. Aug 3, 2020 at 10:06
• @pisco OK thank you Aug 3, 2020 at 10:11
• @pisco its $erfc$ not $erf$. I think its a little bit different. Aug 3, 2020 at 17:41
• @hasan What happend you lost interest in your proble,1 Aug 5, 2020 at 12:22

If $$g(x)$$ is periodic with a period $$T$$, then $$\int_{k}^{T+k} g(x) dx=\int_{0}^{T} g(x) dx ~~~~~(1)$$ $$I=\int_{0}^{2\pi} \text{Erfc}[\cos(a+t)] dt=\int_{a}^{2\pi+a} \text{Erfc}[\cos x] dx=\int_{0}^{2\pi} \text{Erfc}(\cos x) dx~~~~(2)$$ Next note the property that $$\int_{0}^{2a} f(x) dx=\int_{0}^{a}[ f(x)x+ f(2a-x)] dx~~~~(3).$$ As $$\cos(2\pi-x)=\cos x,$$ we get $$I=2\int_{0}^{\pi} \text{Erf}(\cos x) dx~~~~(4)$$ Again using (3), we get $$I=2[\int_{0}^{\pi/2}[\text{Erfc}(\cos x)+ \text{Erfc} (-\cos x)]dx=2\pi,~~~~(5)$$ as $$\text{Erfc}(z)+\text{Erfc}(-z)=2$$.
Edit: Now we take up the pther integral $$J=\int_{0}^{2\pi} [\text{Erfc}(a+\sin x) ~\text{Erfc}(a+\cos x)] dx$$ Due to the property (1) again, we get $$J$$ independent of $$a$$ $$J=\int_{0}^{2\pi} [\text{Erfc}(\sin x) ~\text{Erfc}(\cos x)] dx$$ Using (3) again, we get $$J=\int_{0}^{\pi} [[\text{Erfc}(\sin x) ~\text{Erfc}(\cos x)+[\text{Erfc}(-\sin x) ~\text{Erfc}(\cos x)] dx$$ Next, use $$\text{Erf}(-z)=2-\text{Erf}(z)$$, to write from (4) and (5) $$J=2\int_{0}^{\pi} \text{Erf}(\cos x) dx= I=2\pi$$
• Can you please double check the property you used in equation 3. And can I follow the same policy for $\int\limits_0^{2\pi } {erfc\left( {\cos \left( {a + \theta } \right)} \right)erfc\left( {\sin \left( {a + \theta } \right)} \right)d\theta }$. Thank you. Aug 3, 2020 at 17:46
• Zafar Can you you give me hint to integrate $\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)\operatorname{erfc}\left( {\sin \left( {a + \theta } \right)} \right)d\theta }$ like your reduced form in solution from in equation (3)? It looks so classy approach but I dont know how to start as $sin(a+\theta)$ inside the integral. Thank you. Aug 5, 2020 at 12:49
• @Zafar it seems very beautiful approach. But Can you tell me why did you wrote $(a+sin(x))$ instead of $sin(x+a)$. Aug 5, 2020 at 21:36