0
$\begingroup$

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

enter image description here

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?
$\endgroup$
4
  • 3
    $\begingroup$ 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. $\endgroup$
    – pisco
    Aug 3, 2020 at 10:06
  • $\begingroup$ @pisco OK thank you $\endgroup$ Aug 3, 2020 at 10:11
  • $\begingroup$ @pisco its $erfc$ not $erf$. I think its a little bit different. $\endgroup$
    – hasan
    Aug 3, 2020 at 17:41
  • $\begingroup$ @hasan What happend you lost interest in your proble,1 $\endgroup$
    – Z Ahmed
    Aug 5, 2020 at 12:22

1 Answer 1

3
$\begingroup$

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) $$J=2\int_{0}^{\pi} \text{Erf}(\cos x) dx= I=2\pi$$

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – hasan
    Aug 3, 2020 at 17:46
  • $\begingroup$ 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. $\endgroup$
    – hasan
    Aug 5, 2020 at 12:49
  • $\begingroup$ @hasan OK I will try this, in the meanwhile you may accept this solution with an upvote.. You may put even this second question here in MSE. $\endgroup$
    – Z Ahmed
    Aug 5, 2020 at 12:52
  • $\begingroup$ @Zafar I have put the question in MSE. Can you please take a look? Thank you. $\endgroup$
    – hasan
    Aug 5, 2020 at 13:31
  • $\begingroup$ @Zafar it seems very beautiful approach. But Can you tell me why did you wrote $(a+sin(x))$ instead of $sin(x+a)$. $\endgroup$
    – hasan
    Aug 5, 2020 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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