Difficult integral #2

Lets start by substituting $\sqrt x =u$ and finding the antiderivative The integral becomes $\displaystyle \int \frac{2u}{1+u^{3}} du$

Note that $\left(1+u\right)\left(1-u+u^{2}\right)=1+u^{3}$ ,which gives the integral as $\displaystyle \int \frac{2u}{\left(1+u\right)\left(1-u+u^{2}\right)} du$

Now $\displaystyle\frac{1}{1-u+u^{2}}-\frac{1}{1+u} = \frac{2u-u^{2}}{\left(1+u\right)\left(1-u+u^{2}\right)} =\frac{2u}{\left(1+u\right)\left(1-u+u^{2}\right)}-\frac{u^{2}}{\left(1+u\right)\left(1-u+u^{2}\right)} =\frac{2u}{1+u^{3}} -\frac{u^{2}}{1+u^{3}}$

We have $\displaystyle\frac{1}{1-u+u^{2}}-\frac{1}{1+u} =\frac{2u}{1+u^{3}} -\frac{u^{2}}{1+u^{3}}$

$\displaystyle\frac{1}{1-u+u^{2}}-\frac{1}{1+u}+\frac{u^{2}}{1+u^{3}} =\frac{2u}{1+u^{3}}$

$\displaystyle\int \left( \frac{1}{1-u+u^{2}}-\frac{1}{1+u}+\frac{u^{2}}{1+u^{3}}\right) du =\int \frac{2u}{1+u^{3}} du$

We can find anti-derivative for each one separately, the second and third ones are :

$\displaystyle \int \frac{1}{1+u} du =\ln\left(1+u\right)$

$\displaystyle \int \frac{u^{2}}{1+u^{3}} du =\frac{\ln\left(1+u^{3}\right)}{3}$

Now $\displaystyle 1-u+u^{2} =\left(u-\frac{1}{2}\right)^{2} +\frac{3}{4}$

$\displaystyle\int \frac{1}{1-u+u^{2}}=\displaystyle\int \frac{1}{\left(u-\frac{1}{2}\right)^{2} +\frac{3}{4}} du =\displaystyle \int\frac{4}{3\left( 1+\left(\frac{2u-1}{\sqrt3}\right)^{2}\right)} du$ , by substituting $\frac{2u-1}{\sqrt3}=\tan \theta$ we can find the antiderivative.

The final form of the antiderivative after reverting to original variable and simplifying will be $\displaystyle \frac{2 \arctan \left(\frac{2\sqrt x -1}{\sqrt3}\right)}{\sqrt3} +\ln\left(\frac{\left(1+ x^{\frac{3}{2}}\right)^{\frac{1}{3}}}{1+\sqrt x}\right)$

For the $x=\infty$ limit we can factor out the $\sqrt x$ inside logarithm and in both limits expression inside logarithm evaluates to 1, which evaluates to 0. Evaluating $\displaystyle \frac{2 \arctan \left(\frac{2\sqrt x -1}{\sqrt3}\right)}{\sqrt3}$ at limits will give $\displaystyle \frac{\pi}{\sqrt3} +\frac{\pi}{3\sqrt 3}$ which simplifies to $\frac{4\pi}{3\sqrt 3} =\displaystyle \frac{4\pi}{3^{\frac{3}{2}}}$

The subsitution $u = x^{\frac32}$ gives $\begin{aligned} \int_0^\infty \frac{1}{x^{\frac32}+1}\,dx & = \; \tfrac23\int_0^\infty \frac{u^{-\frac13}}{u+1}\,du \; = \; \tfrac23B(\tfrac13,\tfrac23) \; = \; \tfrac23\Gamma(\tfrac13)\Gamma(\tfrac23) \; = \; \tfrac23 \times\frac{\pi}{\sin\tfrac13\pi} \; = \; \frac{4\pi}{3\sqrt{3}} \; = \; \frac{4\pi}{3^{\frac32}} \end{aligned}$ making the answer $4+3+3+2=\boxed{12}$ .

My solution is based on Contour integral. First, we do a substitution $u=\sqrt x$ such that the integral becomes

$I=\int_0^\infty \dfrac{1}{x^{3/2}+1}dx=\int_0^\infty \dfrac{2u}{u^3+1}du$

Let $\rho$ be a very small positive number and $R$ a very large positive number, $\rho< R$ . Let $C$ be the positively oriented contour consists of four parts: $C=C_1\cup C_2\cup C_3\cup C_4$ , where $C_1=\{x\in\mathbb R| \rho\leq x\leq R\}, C_2=\{Re^{i\theta}| \theta\in[0,2\pi/3]\}, C_3=\{xe^{2\pi i/3}| \rho\leq x\leq R\}, C_4=\{\rho e^{i\theta}| \theta\in[0,2\pi/3]\}.$

Since the function $2u/(u^3+1)$ only has one simple pole at $u=e^{i\pi/3}$ , by residue theorem, the contour integral is $I_C=\int_C \dfrac{2u}{u^3+1}du=2\pi i\underset{u=e^{\pi i/3}}{\operatorname{Res}}\dfrac{2u}{u^3+1}=\dfrac{4\pi}{3}e^{\pi i/6}.$

Taking $\rho \to 0^+$ and $R\to \infty$ , the paths $C_2, C_4$ converges to zero, hence $\begin{aligned} I_C&=\int_{C_1}\dfrac{2udu}{u^3+1}-\int_{-C_3}\dfrac{2udu}{u^3+1}\\ &=\int_0^\infty \dfrac{2udu}{u^3+1}-\int_0^\infty \dfrac{2e^{2\pi i/3}u}{u^3+1}\cdot e^{2\pi i/3}du\text{ (Crucial role of }C_3\text{ here)}\\ &=(1-e^{4\pi i/3})I\\ &=\sqrt3 e^{\pi i/6}I.\end{aligned}$ Comparing two expression of $I_C$ we see $I=\boxed{\dfrac{4\pi }{3\sqrt 3}}.$