# Verification of an Cauchy's contour Integral of Complementary Error function?

I tried to find an integral of the following,$$\DeclareMathOperator{\erfc}{erfc}$$

$$\int\limits_0^{2\pi} \erfc(a + b\cos(\theta))\erfc(c + d\sin(\theta))\,d\theta$$

Where, $$a,b,c,d \in \Bbb R$$

Now, $$\cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2};\quad\sin(\theta) = \frac{e^{j\theta } - e^{-j\theta }}{2j}$$

Let $$z = {e^{j\theta }}$$

Therefore, I can rewrite the integral as, $$\oint\limits_{|z| = 1} {\erfc\left( {a + b\left( {\frac{{z + {z^*}}}{2}} \right)} \right)erfc\left( {c + d\left( {\frac{{z - {z^*}}}{{2j}}} \right)} \right)\frac{{dz}}{{iz}}}$$

Again, let $$f(z) = \erfc\left(a + b\left(\frac{z + z^*}{2}\right)\right)\erfc\left(c + d\left(\frac{z - z^*}{2j}\right) \right)$$

Hence , the final integral can be written as,

$$\oint\limits_{|z| = 1}f(z)\frac{dz}{iz} = 2\pi f(0) = 2\pi\erfc(a)\erfc(c)$$

As $$0$$ falls inside $$|z| = 1$$.

Can you tell me if I am correct or wrong or it needs more reasoning? Thank you.

• the $z^\ast$ in your integrand is $1/z$ on the unit circle, so you have poles at the origin, the integrand is not analytic inside the unit circle; and indeed, the answer is incorrect, for example, for $a=b=c=d=1$ the integral equals $0.6268$ while $2\pi \,\text{erfc}(1)\text{erfc}(1)=0.1555.$ --- also crossposted at math.stackexchange.com/q/3781857/87355 – Carlo Beenakker Aug 7 at 9:47
• @CarloBeenakker does that mean the integration is not possible in close from? Can you give me a hint to do that. Yes I posted this in overflow as no one answered in MSE Thank you. – hasan Aug 7 at 10:02
• indeed, no closed form expression as far as I can see. – Carlo Beenakker Aug 7 at 10:05
• @CarloBeenakker I think I am wrong. It depends on $a,b,c,d$ – hasan Aug 7 at 10:22