Can anyone help with solving these two definite integrals using contour integration:
$$\int_{0}^{1}\frac{\sqrt{x(1-x)}}{(2x+1)^{2}(x+2)}dx$$
$$\int_{1}^{2}\frac{dx}{x\sqrt[7]{(x-1)^{3}(2-x)^{4}}}$$.
For the first one, I use a dog bone contour around $[0,1]$ on the real axis, and I have two poles at $-1/2$ and $-2$, the first one of the second order and the second is simple. At infinity the integral over the big circle is zero as R tends to infinity (Jordan's lemma). When I calculate the residues at the poles, I get $-\frac{2\sqrt{3}i}{3}$ for $-1/2$ and I believe I should get $-\frac{\sqrt{3}i}{6}$, and for the second pole I get $\frac{\sqrt{6}i}{9}$, which I think is correct. Can anyone help me with finding the mistake in calculating the residue at the pole $-1/2$?
Similarly, for the second integral, I use a dog bone contour around $[1,2]$ on the real axis, and have only one pole at 0, so I believe that to get the solution I only have to calculate the residue at the pole 0, and I can not get the right solution.
Can someone help me?
For the second one, use a dogbone contour traversed counterclockwise around $z=1$ and $z=2$ with the function
$$f(z) = \frac{1}{z} \exp((-3/7)\mathrm{LogA}(z-1)) \exp((-4/7)\mathrm{LogB}(2-z))$$
with $\mathrm{LogA}$ having argument between $[0,2\pi)$ and branch cut on the positive real axis and $\mathrm{LogB}$ having argument between $[-\pi, \pi)$ and branch cut on the negative real axis. We claim the branch cut of $f(z)$ is the interval $[1,2]$ with continuity across the cut on $(2,\infty)$ (and hence analyticity by Morera's theorem).
For the first interval put $x=1+t+i\epsilon$ with $\epsilon\ge 0$ and $t\in(0,1)$ so we are above the cut from the first logarithm. We are in the half plane not containing the cut of the second logarithm. We obtain
$$\frac{1}{x} \exp((-3/7)\log(x-1)) \exp((-4/7)\log(2-x)) \\ = \frac{1}{x\sqrt[7]{(x-1)^3(2-x)^4}}.$$
Next put $x=1+t-i\epsilon.$ We are now below the cut from the first logarithm. The cut of the second logarithm does not participate. We get
$$\frac{1}{x} \exp((-3/7)\log(x-1)+(-3/7)\times 2\pi i) \exp((-4/7)\log(2-x)) \\ = \frac{\exp((-6/7)\pi i)}{x\sqrt[7]{(x-1)^3(2-x)^4}}.$$
The conclusion is that with
$$J = \int_1^2 \frac{1}{x\sqrt[7]{(x-1)^3(2-x)^4}} \; dx$$
the two straight segments from the dogbone contribute
$$(-1 + \exp((-6/7)\pi i)) J.$$
To see that we have continutity across $(2,\infty)$ we first put $x = 2 + t + i\epsilon$ so we are above the cut of $\mathrm{LogA}.$ We are now below the cut of $\mathrm{LogB}.$ Combine to get
$$\frac{1}{x} \exp((-3/7)\log(x-1)) \exp((-4/7)\log(x-2)+(-4/7)\times -\pi i) \\ = \frac{\exp((4/7)\pi i)}{x\sqrt[7]{(x-1)^3(x-2)^4}}.$$
Next put $x = 2 + t -i\epsilon$ so we are below the cut of $\mathrm{LogA}.$ We are now above the cut of $\mathrm{LogB}.$ Combine to get
$$\frac{1}{x} \exp((-3/7)\log(x-1)+(-3/7)\times 2\pi i) \exp((-4/7)\log(x-2)+(-4/7)\times \pi i) \\ = \frac{\exp((-10/7)\pi i)}{x\sqrt[7]{(x-1)^3(x-2)^4}}.$$
Seeing that $\exp((-10/7)\pi i) = \exp((4/7)\pi i)$ we indeed have continuity across the cut as claimed. Observe also that $\lim_{R\to\infty} 2\pi R \times 1/R/\sqrt[7]{R^3\times R^4} = 0$ so the residue at infinity is zero. This leaves the residue at zero which is particularly simple, we have
$$\left.\exp((-3/7)\mathrm{LogA}(z-1)) \exp((-4/7)\mathrm{LogB}(2-z))\right|_{z=0} \\ = \exp((-3/7)\mathrm{LogA}(-1)) \exp((-4/7)\mathrm{LogB}(2)) = \exp((-3/7)\pi i) 2^{-4/7}.$$
We then have for our integral that
$$(-1 + \exp((-6/7)\pi i)) J = - 2\pi i \exp((-3/7)\pi i) 2^{-4/7}$$
so that
$$J = - 2\pi i \frac{\exp((-3/7)\pi i)}{-1+\exp((-6/7)\pi i)} 2^{-4/7} \\ = - 2\pi i \frac{1}{-\exp((3/7)\pi i)+\exp((-3/7)\pi i)} 2^{-4/7}.$$
Hence the desired answer is
$$\bbox[5px,border:2px solid #00A000]{ J = \frac{\pi}{\sin(3\pi/7)} 2^{-4/7}.}$$
For the two circular components of the dogbone we note that $\lim_{\epsilon\to 0} 2\pi \epsilon \times 1/1/\epsilon^{3/7}/1^{4/7} = 0$ and $\lim_{\epsilon\to 0} 2\pi \epsilon \times 1/2/1^{3/7}/\epsilon^{4/7} = 0$ so these two really do vanish. (Parameterize by $1+\epsilon \exp(i\theta)$ and $2+\epsilon\exp(i\theta).$)