Find using Residue Theorem the following integral

Question

Find using Residue Theorem

$$\int_{-1}^1 \dfrac{dx}{(\sqrt {1-x^2})(1+x^2)}$$

My try:

I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.

and $C_2$ is the line joining $-1$ to $1$.

I now consider the function in $\Bbb C$ to be $$\int_C \dfrac{dz}{(\sqrt {1-z^2})(1+z^2)}$$

Over $C_2$ I get that $$\int_{C_2} \dfrac{dz}{(\sqrt {1-z^2})(1+z^2)}$$=$$\int_{-1}^1 \dfrac{dx}{(\sqrt {1-x^2})(1+x^2)}$$

But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.

Also I cant figure out how to calculate $$\int_{C_1} \dfrac{dz}{(\sqrt {1-z^2})(1+z^2)}$$

Any help from someone here?

Thanks for reading my post

Answer 1

The complex function has two branch points at $z=\pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.

Let's pick a contour similar to the one seen in this example.

We pick the branch cut on $[-1,1]$ on the real line such that

\begin{align} 1+z &= r_1e^{i\phi_1}, & \phi_1 \in (-\pi,\pi] \\ 1-z &= r_2e^{i\phi_2}, & \phi_2 \in (0,2\pi) \end{align}

The contour consists of:

• $C_1$: a line segment just above the branch cut going from $1+\epsilon$ to $-1+\epsilon$,
• $C_2$: a left semicircle $z = -1 +\epsilon e^{it}$, going counter-clockwise from $-1+\epsilon$ to $-1-\epsilon$
• $C_3$: a line segment just below the branch cut, going from $-1-\epsilon$ to $1-\epsilon$
• $C_4$: a right semicircle $z = 1 + \epsilon e^{it}$, going counterclockwise from $1-\epsilon$ to $1+\epsilon$

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In the limit of $\epsilon \to 0$, we have

$$ f(z)\big|_{C_1} = \frac{1}{(z^2+1)\sqrt{|1+z|}e^{i0/2}\sqrt{|1-z|}e^{i2\pi/2}} = \frac{-1}{(z^2+1)\sqrt{|1+z|}\sqrt{|1-z|}} $$

$$ f(z)\big|_{C_3} = \frac{1}{(z^2+1)\sqrt{|1+z|}e^{-i0/2}\sqrt{|1-z|}e^{i0/2}} = \frac{1}{(z^2+1)\sqrt{|1+z|}\sqrt{|1-z|}} $$

$$ \implies \int_{C_1} f(z)\ dz + \int_{C_3} f(z)\ dz = -\int_1^{-1} f(x) dx + \int_{-1}^1 f(x) dx = 2\int_{-1}^1 f(x)\ dx $$

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Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality

$$ \int_{C_2} f(z)\ dz \le \frac{L(C_2)}{|1+z^2|\sqrt{|1-z|}\sqrt{|1+z|}} \le \frac{\pi \epsilon}{2\sqrt{2}\sqrt{\epsilon}} \to 0 $$

Since $|z| \ge 1$ and $|1-z| \ge 2$

$$ \int_{C_4} f(z)\ dz \le \frac{L(C_4)}{|1+z^2|\sqrt{|1-z|}\sqrt{|1+z|}} \le \frac{\pi \epsilon}{2\sqrt{2}\sqrt{\epsilon}} \to 0 $$

Since $|z| \ge 1$ and $|1+z| \ge 2$

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Finally, use residues to finish the rest. You may also need to find the residue at infinity.