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 } $
- Is it not possible to find the closed form of the integral of complimentary error function with trigonometric function inside.
- Can anyone share me any integral of complimentary error function that has trigonometric function inside as argument?
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) $$J=2\int_{0}^{\pi} \text{Erf}(\cos x) dx= I=2\pi$$