How Can I Find a Symbolic Solution for This Complex Integral?

[Zoran]

Hi All: I am working on a research project and am trying to get a symbolic solution to the following integral:

$$\int_{0}^{\frac{\pi }{4}}\int_{0}^{\frac{\pi }{4}}\frac{\cos\left ( z_{1} \right )\cos \left ( z_{2} \right )}{\sqrt{\left ( \cos\left ( z_{1} \right ) ^{2}+\frac{1}{2} \right )\left ( \cos\left ( z_{2} \right ) ^{2}+\frac{1}{2} \right )-\frac{1}{4}}}dz_{1}dz_{2}$$

The problem seems to be with the inclusion of the -1/4 in the denominator i.e. without it the integral has the solution:

$$\arctan\left ( \frac{1}{\sqrt{2}} \right ) ^{2}$$
.
Any help is appreciated.

 

[jvicens]

Hello! Thank you for sharing your research project with us. I understand that you are trying to find a symbolic solution to the integral with the added -1/4 in the denominator. This can certainly be a challenging task, but I will try my best to provide some guidance.

Firstly, let's break down the integral into two separate integrals, one for each variable z1 and z2. This will make it easier to handle the complex denominator. We can also use the trigonometric identity cos^2(x) = 1/2 + 1/2cos(2x) to simplify the denominator.

So, our integral becomes:

\int_{0}^{\frac{\pi }{4}} \frac{\cos(z_{1})}{\sqrt{\left ( \frac{1}{2}+\frac{1}{2}\cos(2z_{1}) \right )\left ( \cos(z_{2})^{2}+\frac{1}{2} \right )-\frac{1}{4}}} dz_{1} * \int_{0}^{\frac{\pi }{4}} \frac{\cos(z_{2})}{\sqrt{\left ( \frac{1}{2}+\frac{1}{2}\cos(2z_{2}) \right )\left ( \cos(z_{1})^{2}+\frac{1}{2} \right )-\frac{1}{4}}} dz_{2}

Now, we can use the substitution u = cos(z1) and v = cos(z2) to simplify the integrals further. This will result in:

\int_{\frac{\sqrt{2}}{2}}^{1} \frac{du}{\sqrt{\left ( \frac{1}{2}+\frac{1}{2}u^{2} \right )\left ( v^{2}+\frac{1}{2} \right )-\frac{1}{4}}} * \int_{\frac{\sqrt{2}}{2}}^{1} \frac{dv}{\sqrt{\left ( \frac{1}{2}+\frac{1}{2}v^{2} \right )\left ( u^{2}+\frac{1}{2} \right )-\frac{1}{4}}}

Now, we can use a partial fraction decomposition to separate the fractions in the denominators. This will