Evaluate
$$I=\int_{-1}^{1} \frac{1}{(1+x^4)\sqrt{1-x^2}}dx$$
by integrating along the following complex contour:
When we take $R\rightarrow 1$ and $ε\rightarrow 0$, the contour integral itself will be $2πi$ times the sum of the 4 residues of the function, which we can get easily enough using L'Hopital's rule, or factorising $1+x^4$ and so on. My real trouble comes when we consider the lower integral along the real line,
$$\int_{1}^{-1} \frac{1}{(1+x^4)\sqrt{1-x^2}}=\int_{1}^{-1} \frac{1}{(1+x^4)\sqrt{1-xe^{2πi}}\sqrt{1+xe^{2πi}}}$$
Where I have included $e^{2πi}$ explicitly since it's the lower path. My intuition is telling me that this piece of the contour shouldn't just be the same as the piece above because of this $e^{2πi}$ inside the square root. Furthermore, if it were the exact same then the two pieces would cancel exactly…
So how we can we separate out the 'effects' of this $e^{2πi}$ to get this piece in the form of constant * $I$?
Best Answer
To evaluate this integral by means of the residue theorem, we have to create a closed contour, inside which the integrand is a single-valued function. For this purpose we also make necessary cuts. In your case the best option is the usage of a dogbone contour with the cut from $-1$ to $1$, but for the sake of simplicity we can preliminary modify the integral a bit. $$I=\int_{-1}^{1} \frac{1}{(1+t^4)\sqrt{1-t^2}}dt\overset{t^2=x}{=}\int_0^1\frac1{1+x^2}\frac{dx}{\sqrt{x(1-x)}}$$ This substitution will allow us to evaluate only two residues (at $x=\pm i$) instead of four of the initial integral.
We also note that the integral along the dogbone contour (clockwise) with the cut $[0;1]$ gives us $2I$ (the integrals along the arches around $x=0$ and $x=$ tend to zero). Now we consider the integral along the following closed contour $C$:
We added a big circle with the radius $R\to\infty$ and a path $1$, which links the dogbone contour with this circle. Inside the contour the function is single-valued.
Therefore, $$\oint_C\frac1{1+x^2}\frac{dx}{\sqrt{x(1-x)}}=2I+I_1-I_1+I_R=2\pi i\underset{x=e^{\frac{\pi i}2}; \,e^{\frac{3\pi i}2}}{\operatorname{Res}}\frac1{1+x^2}\frac1{\sqrt{x(1-x)}}$$ The integral $I_R\to 0$ as $R\to\infty$, and integrals along the path $1$ cancel each other (we integrate twice, in the opposite directions). $$I=\pi i\underset{x=e^{\frac{\pi i}2}; \,e^{\frac{3\pi i}2}}{\operatorname{Res}}\frac1{(x+i)(x-i)}\frac1{\sqrt{x(1-x)}}=\pi i\left(\frac1{2i}\frac1{\sqrt{e^{\frac{\pi i}2}(1-e^{\frac{\pi i}2})}}-\frac1{2i}\frac1{\sqrt{e^{\frac{3\pi i}2}(1-e^{\frac{3\pi i}2})}}\right)$$ $$=\frac\pi2\left(e^{-\frac{\pi i}4}(1-i)^{-1/2}+e^{\frac{\pi i}4}(1+i)^{-1/2}\right)=\frac\pi{\sqrt2}\Re \,e^{-\frac{\pi i}4}\sqrt{1+i}$$ $$\boxed{\,\,I=\frac\pi22^\frac34\cos\frac\pi8=\frac\pi2\sqrt{1+\sqrt2}\,\,}$$