# Find using Residue Theorem the following integral

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

• Either all the time there is lacking some parentheses in the denominator or else there is rather weird $\;1\;$ there... – DonAntonio Dec 4 '18 at 17:02
• @DonAntonio;edited my question,can u have a look now – user596656 Dec 4 '18 at 17:07
• Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $\;z=-1;$ to the point $\;z=1\;$ since on those points your function isn't defined. You can't also go through the point $\;z=i\;$ for the same reason. – DonAntonio Dec 4 '18 at 18:14

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 $$1+\epsilon$$

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

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

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

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

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

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

• Thanks for an answer,I will need some time and patience to understand it – user596656 Dec 5 '18 at 10:09