Real part of the integral $\int_0^{ \infty } \frac{1}{\sqrt{1-x^2+\sqrt{1-x^2}+1}}dx$

Question

I'm trying to calculate real part of this integral $\int_0^{ \infty } \frac{1}{\sqrt{1-x^2+\sqrt{1-x^2}+1}}dx$. Is there a way to calculate using residue theorem? Thank you.

Answer 1

This is not really what the OP wants (since the real part is supposed to equal $\pi/2$, which is quite hard to guess from my results), but I believe the following may be useful for a more general case.

If we want to separate real and imaginary parts of the integral, we can do it directly:

$$I = \int_0^{ \infty } \frac{1}{\sqrt{2-x^2+\sqrt{1-x^2}}}dx= I_1+I_2= \\ =\int_0^1 \frac{1}{\sqrt{2-x^2+\sqrt{1-x^2}}}dx+\int_1^{ \infty } \frac{1}{\sqrt{2-x^2+\sqrt{1-x^2}}}dx$$

$I_1$ is real and can be found by usual real methods.

$I_2$ needs to be transformed to separate the real part. Let, for simplicity $x=1+y$, then:

$$I_2=\int_0^{ \infty } \frac{1}{\sqrt{1-y(y+2)+i\sqrt{y(y+2)}}}dy=\int_0^{ \infty } \frac{\sqrt{1-y(y+2)-i\sqrt{y(y+2)}}}{\sqrt{(1-y(y+2))^2+y(y+2)}}dy$$

In the denominator we obtain:

$$y^4 + 4 y^3 + 3 y^2 - 2 y + 1>0$$

So we only need to separate the real and imaginary parts in the numerator:

$$\sqrt{a-ib}=(a-ib)^{1/2}=\left( \sqrt{a^2+b^2}e^{-i \arctan (b/a)} \right)^{1/2}=\sqrt[4]{a^2+b^2} \exp \left(-i \frac{ \arctan (b/a)}{2} \right)$$

The real part of this expression is:

$$\sqrt[4]{a^2+b^2} \cos \left(\frac{ \arctan (b/a)}{2} \right)=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}=\frac{1}{\sqrt{2}}\sqrt{a+\sqrt{a^2+b^2}}$$

So the real part of $I_2$ should be:

$$\Re (I_2)=\frac{1}{\sqrt{2}} \int_0^{ \infty } \frac{\sqrt{1-y(y+2)+\sqrt{y^4 + 4 y^3 + 3 y^2 - 2 y + 1}}}{\sqrt{y^4 + 4 y^3 + 3 y^2 - 2 y + 1}}dy$$

Now we have:

$$\Re (I)=I_1+\Re (I_2)$$

The integrals (especially the second one) don't look nice, but they are perfectly real.

Also, for large $y$ the integrand falls like $1/y^2$, which confirms that this integral converges.

With the help of Wolfram Alpha we have:

$$I_1 = 0.6524026393603789453511342861873757483364 \ldots$$

$$\Re (I_2)=\frac{1}{\sqrt{2}}1.2988048083677320189625984238871380167 \ldots$$

The result checks our numerically, as:

$$I_1+\Re (I_2)= \frac{\pi}{2}$$