How do I integrate an erfc (complementary error function)?

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Svante Monie
Svante Monie 2015-7-15
编辑： Svante Monie 2015-7-15
I'm trying to calculate following in Matlab:
U = 1/pi * integral from [0, 1/Tao] {erfc(gamma*f(x)/sqrt(Tao - f(x))} dx where f(x) = 3 * sin(x) - x * cos(x)/(sin(x))^3 gamma, Tao = variables
I've tried many different ways, but none works. The last attempt looks like this:
gamma = 0.7993;
TAO = [0,0.6127,1.2255,1.8382,2.4509,3.0636,3.6764,4.2891,4.9018,5.5145,6.1273];
f0TAO = zeros();
for i=2:length(TAO) f0TAO(i) = (3*((sin(TAO(i))-(TAO(i)*cos(TAO(i))))/(sin(TAO(i)))^3));
end
f0Tao = 1./f0TAO;
Q = zeros();
for i=2:length(Tao);
c = TAO(i);
f = erfc(gamma.*((3.*((sin(x)-(x.*cos(x)))./(sin(x))^3))./sqrt(c-(3.*((sin(x)-(x.*cos(x)))./(sin(x))^3)))));
Q(i) = 1/pi*int(f,0,f0Tao(i));
end
This resulted in an error like this:
The following error occurred converting from sym to double: Error using mupadmex Error in MuPAD command: DOUBLE cannot convert the input expression into a double array.
If the input expression contains a symbolic variable, use the VPA function instead.
What should I do? Can anyone give me some brilliant clues?
Thanks in advance!

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Torsten
Torsten 2015-7-15
Q(i)=1/pi*vpa(int(f,0,f0Tao(i)),5);
But I guess the integration will be difficult because you divide by sin(x)^3 which is zero at x=0.
Additionally, the sqrt will become complex-valued.
Best wishes
Torsten.
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Torsten
Torsten 2015-7-15
If
f=@(x)erfc(gamma.*((3.*((sin(x)-(x.*cos(x)))./(sin(x))^3))./sqrt(c-(3.*((sin(x)-(x.*cos(x)))./(sin(x))^3)))));
f(2);
still gives an error, check the value of the argument to the erfc function and whether gamma and c are real and not symbolic.
Best wishes
Torsten.
Svante Monie
Svante Monie 2015-7-15
编辑：Svante Monie 2015-7-15
Now I have found the cause of the error. In the denominator I have sqrt(c-(3.*((sin..., this gives complex values and thats what MatLab don't like. It needs _real_ input, as stated. When I put abs() around the expression under the sqrt, erfc delievers! Question is then of course, if the integral is valid for my intentions...

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