Solve Arctan(x)/sqrt(x^3) Integral With Series Or Complex Analysis

What's up, calculus crew! Today, we're diving deep into the fascinating world of improper integrals, specifically tackling this gnarly beast:

$$\int_{0}^{+\infty} \frac{\arctan x}{\sqrt{x^3}} \, dx$$

This bad boy looks intimidating, right? But don't sweat it, guys. We're going to break it down using some seriously cool mathematical tools: series expansions and complex analysis. Whether you're a seasoned mathlete or just dipping your toes into advanced calculus, stick around, because this is gonna be a wild ride.

The Taylor Series Approach: Unpacking Arctan(x)

First off, let's talk about that $\arctan x$ in the numerator. You know, the inverse tangent function? It's a total rockstar when it comes to Taylor series expansions. Remember this gem?

$$\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dotsb$$

This expansion is valid for $|x| \leq 1$. Now, our integral goes all the way to infinity, so we need to be a bit careful. But hey, we can at least start by plugging this series into our integral and see what happens. Let's rewrite the integral with the series:

$$\int_{0}^{\infty} \frac{1}{\sqrt{x^3}} \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \right) \, dx$$

We can swap the integral and the summation (under certain conditions, of course – gotta keep our math honest!):

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{\infty} x^{2n+1 - \frac{3}{2}} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{\infty} x^{2n - \frac{1}{2}} \, dx$$

Now, let's evaluate that inner integral. We're looking at $\int_{0}^{\infty} x^p \, dx$, where $p = 2n - \frac{1}{2}$. This integral only converges if $p < -1$. So, we need $2n - \frac{1}{2} < -1$, which means $2n < -\frac{1}{2}$, or $n < -\frac{1}{4}$. Uh oh. This means the Taylor series expansion around $x=0$ alone isn't going to cut it for the entire range of integration from $0$ to $\infty$. The function behaves differently for large $x$.

So, what's the deal? When $|x| > 1$, the Taylor series for $\arctan x$ actually diverges. We need another trick up our sleeve. This is where things get really interesting, and why we need to consider the behavior of $\arctan x$ for larger values of $x$. As $x \to \infty$, $\arctan x$ approaches $\frac{\pi}{2}$. This limit is crucial! Our integral is essentially a weighted average of $\arctan x$ over the positive real axis. The $\frac{1}{\sqrt{x^3}}$ term acts as a decaying weight, meaning the behavior of $\arctan x$ near $0$ and its limiting value at infinity both play significant roles. The divergence issue with the standard Taylor series at larger $x$ signals that we can't simply term-by-term integrate an expansion valid only near zero. We need a method that can handle the function's behavior across its entire domain of integration, or at least handle the problematic parts separately. This is a common pitfall when working with improper integrals and series – always check the convergence criteria across the entire interval!

The Complex Analysis Conundrum: Winding Our Way Through

Alright, since the series approach hit a snag for large $x$, let's bring out the heavy artillery: complex analysis. This is where we transform our real integral into a contour integral in the complex plane. It sounds fancy, and frankly, it is pretty slick.

We're going to consider the integral:

$$\oint_C \frac{\log z}{\sqrt{z^3}} \, dz$$

Why $\log z$ instead of $\arctan z$? Well, $\arctan z$ has a logarithmic representation: $\arctan z = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right)$. Using the logarithm directly in the complex plane is often more straightforward for contour integration. We'll use a keyhole contour to handle the branch cut of $\sqrt{z}$ and $\log z$, which typically lies along the positive real axis. This contour consists of a large circle, a small circle around the origin, and two line segments just above and below the positive real axis.

Let's define our function $f(z) = \frac{\log z}{z^{3/2}}$. We need to be careful about the branch of $z^{3/2}$. We'll choose the principal branch where $z^{3/2} = e^{\frac{3}{2} \log z}$, with $\text{Arg}(z) \in (-\pi, \pi]$. This means $\log z = \ln|z| + i \text{Arg}(z)$.

Our contour $C$ will be made up of four parts:

1. $C_R$: A large circle of radius $R$ centered at the origin. As $R \to \infty$, the integral over $C_R$ will go to zero because the integrand decays sufficiently fast. For $z$ on $C_R$, $|z|=R$, so $|z^{3/2}| = R^{3/2}$. $|\log z| \approx \ln R$. Thus, $\left| \int_{C_R} f(z) dz \right| \leq 2\pi R \cdot \frac{\ln R}{R^{3/2}} \to 0$ as $R \to \infty$.

2. $C_{\epsilon}$: A small circle of radius $\epsilon$ centered at the origin. As $\epsilon \to 0$, the integral over $C_{\epsilon}$ will also go to zero. For $z$ on $C_{\epsilon}$, $|z|=\epsilon$, so $|z^{3/2}| = \epsilon^{3/2}$. $|\log z| \approx \ln \epsilon$. Thus, $\left| \int_{C_{\epsilon}} f(z) dz \right| \leq 2\pi \epsilon \cdot \frac{|\ln \epsilon|}{\epsilon^{3/2}} = 2\pi \frac{|\ln \epsilon|}{\sqrt{\epsilon}} \to 0$ as $\epsilon \to 0$.

3. $L_1$: The line segment just above the positive real axis from $\epsilon$ to $R$. Here, $z = x$, where $x$ is real and positive. We use $\log z = \ln x + i \cdot 0 = \ln x$ (since we're just above the axis, Arg(z) is close to 0) and $z^{3/2} = x^{3/2}$. So, $\int_{L_1} f(z) dz = \int_{\epsilon}^R \frac{\ln x}{x^{3/2}} \, dx$.

4. $L_2$: The line segment just below the positive real axis from $R$ to $\epsilon$. Here, $z = x e^{-2\pi i}$ (due to the branch cut), so $\log z = \ln x + i (2\pi)$ and $z^{3/2} = x^{3/2} e^{-3\pi i} = -x^{3/2}$. Oh wait, that's not right for the principal branch. For the principal branch with Arg(z) in $(-\pi, \pi]$, when we cross the negative real axis, the argument changes. Let's rethink the contour. A standard keyhole contour cuts along the positive real axis. We approach from above and below.

Let's use a contour that goes around the origin and has a branch cut along the positive real axis. Consider the integral $\oint_C \frac{(\log z)}{z^{3/2}} dz$. The branch cut for $z^{3/2}$ is along the positive real axis. We choose the branch of $\log z$ such that $\log z = \ln r + i\theta$ with $0 \leq \theta < 2\pi$. Then $z^{3/2} = r^{3/2} e^{i 3\theta/2}$.

Our contour $C$ is a large circle $C_R$, a small circle $C_{\epsilon}$, and two segments along the positive real axis. Let the segment above be $L_1$ and the segment below be $L_2$.

On $L_1$, $z=x$, $\theta=0$. $\log z = \ln x$. $z^{3/2} = x^{3/2}$. $\int_{L_1} \frac{\log z}{z^{3/2}} dz = \int_{\epsilon}^R \frac{\ln x}{x^{3/2}} dx$.

On $L_2$, we approach the real axis from just below, so $\theta = 2\pi$. $z=x e^{i 2\pi}$. $\log z = \ln x + i 2\pi$. $z^{3/2} = x^{3/2} e^{i 3(2\pi)/2} = x^{3/2} e^{i 3\pi} = -x^{3/2}$. $\int_{L_2} \frac{\log z}{z^{3/2}} dz = \int_R^{\epsilon} \frac{\ln x + i 2\pi}{-x^{3/2}} dx = -\int_{\epsilon}^R \frac{\ln x + i 2\pi}{-x^{3/2}} dx = \int_{\epsilon}^R \frac{\ln x + i 2\pi}{x^{3/2}} dx$.

Adding these two segments, we get:

$$\int_{L_1} + \int_{L_2} = \int_{\epsilon}^R \frac{\ln x}{x^{3/2}} dx + \int_{\epsilon}^R \frac{\ln x + i 2\pi}{x^{3/2}} dx = 2 \int_{\epsilon}^R \frac{\ln x}{x^{3/2}} dx + i 2\pi \int_{\epsilon}^R \frac{1}{x^{3/2}} dx$$

As $\epsilon \to 0$ and $R \to \infty$, the integrals over $C_R$ and $C_{\epsilon}$ vanish. The sum of the integrals over $L_1$ and $L_2$ becomes:

$$2 \int_{0}^{\infty} \frac{\ln x}{x^{3/2}} dx + i 2\pi \int_{0}^{\infty} x^{-3/2} dx$$

This seems complicated. Let's try a different approach with $\arctan z$. Recall $\arctan z = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right)$. Our integral is $\int_{0}^{\infty} \frac{\arctan x}{x^{3/2}} dx$. Let's use a semicircular contour in the upper half-plane, but we need to avoid the singularities of $\arctan z$. The singularities are where $1-iz=0$ or $1+iz=0$, which are $z=1/i = -i$ and $z=-1/i = i$. So, $z=i$ is inside our upper-half plane semicircle. This makes the semicircle contour tricky.

A Better Complex Analysis Strategy: Let's consider the integral $I = \int_{0}^{\infty} \frac{\arctan x}{x^{3/2}} dx$. We can use a Bromwich contour or a Pochhammer contour, but those are often for specific functions like the Gamma function. Let's stick to a modified keyhole contour that avoids the origin and handles the branch cut appropriately.

Consider the function $f(z) = \frac{\arctan z}{z^{3/2}}$. We'll use a contour that hugs the positive real axis. Let's use the identity $\arctan x = \frac{\pi}{2} - \arctan \frac{1}{x}$ for $x>0$. This doesn't seem to simplify things much for the integration.

Let's go back to $\arctan x = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right)$. Substitute this into the integral:

$$I = \int_0^{\infty} \frac{1}{2i} \frac{\log \left( \frac{1+iz}{1-iz} \right)}{z^{3/2}} dz$$

This still leads to complicated functions. A common technique for integrals involving $\arctan x$ is to use a different contour or a substitution. Let $x = \tan \theta$. Then $dx = \sec^2 \theta \, d\theta$. When $x=0$, $\theta=0$. When $x \to \infty$, $\theta \to \frac{\pi}{2}$.

$$I = \int_0^{\pi/2} \frac{\theta}{(\tan \theta)^{3/2}} \sec^2 \theta \, d\theta = \int_0^{\pi/2} \frac{\theta \cos^{3/2} \theta}{\sin^{3/2} \theta} \frac{1}{\cos^2 \theta} \, d\theta = \int_0^{\pi/2} \frac{\theta}{\sin^{3/2} \theta \cos^{1/2} \theta} \, d\theta$$

This looks like a Beta function integral, but with $\theta$. This substitution transforms the integral into a form that might be solvable using special functions, but it doesn't directly employ complex analysis in the standard way of contour integration of a complex function.

The Gamma Function Connection: A Deeper Dive

Let's try a different approach that bridges calculus and special functions, which often arise from complex integration. Consider the integral $I = \int_0^{\infty} \frac{\arctan x}{x^{3/2}} dx$. We can use integration by parts. Let $u = \arctan x$ and $dv = x^{-3/2} dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = \frac{x^{-1/2}}{-1/2} = -2x^{-1/2}$.

$$I = \left[ -2x^{-1/2} \arctan x \right]_0^{\infty} - \int_0^{\infty} (-2x^{-1/2}) \frac{1}{1+x^2} dx$$

Let's evaluate the boundary terms:

• At the upper limit ($x \to \infty$): $-2x^{-1/2} \arctan x \approx -2 \frac{1}{\sqrt{x}} \cdot \frac{\pi}{2} = -\frac{\pi}{\sqrt{x}} \to 0$.
• At the lower limit ($x \to 0$): $-2x^{-1/2} \arctan x$. Since $\arctan x \approx x$ for small $x$, this term is like $-2x^{-1/2} \cdot x = -2x^{1/2} \to 0$.

So, the boundary terms vanish! This leaves us with:

$$I = 2 \int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$$

Now this integral looks much more manageable using complex analysis! We can evaluate $\int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$ using a semicircular contour in the upper half-plane with a branch cut along the positive real axis for $z^{-1/2}$.

Let $f(z) = \frac{z^{-1/2}}{1+z^2}$. We choose the branch $z^{-1/2} = e^{-\frac{1}{2} \log z}$ with $\text{Arg}(z) \in [0, 2\pi)$. The poles of $f(z)$ are the roots of $1+z^2 = 0$, which are $z = i$ and $z = -i$. Only $z=i$ is inside our upper-half plane contour.

For the pole at $z=i$: $z=i = e^{i\pi/2}$. The residue is

$$\text{Res}(f, i) = \lim_{z \to i} (z-i) f(z) = \lim_{z \to i} (z-i) \frac{z^{-1/2}}{(z-i)(z+i)} = \frac{i^{-1/2}}{2i}$$

Now, we need to evaluate $i^{-1/2}$. Since $i = e^{i\pi/2}$, $i^{-1/2} = (e^{i\pi/2})^{-1/2} = e^{-i\pi/4} = \cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} = \frac{1-i}{\sqrt{2}}$.

So, the residue is:

\text{Res}(f, i) = \frac{\frac{1-i}{\sqrt{2}}}{2i} = \frac{1-i}{2i\sqrt{2}} = \frac{(1-i)(-i)}{2i\sqrt{2}(-i)} = \frac{-i + i^2}{2(-i^2)}\sqrt{2}} = \frac{-1-i}{2\sqrt{2}}

By the Residue Theorem, the integral over the closed contour is $2\pi i$ times the sum of the residues. The integral along the real axis (from $0$ to $\infty$) is $\int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$. The integral over the large semicircle vanishes as its radius goes to infinity. The integral along the negative real axis needs careful handling due to the branch cut.

Let's use the standard keyhole contour along the positive real axis. The function is $g(z) = \frac{z^{-1/2}}{1+z^2}$. Branch cut along positive real axis, $z^{-1/2} = e^{-\frac{1}{2}(\ln r + i\theta)}$, with $\theta \in [0, 2\pi)$. Poles at $z=i$ and $z=-i$.

For $z=i = e^{i\pi/2}$: $\theta=\pi/2$. $i^{-1/2} = e^{-\frac{1}{2}(i\pi/2)} = e^{-i\pi/4}$. Residue at $z=i$ is $\frac{e^{-i\pi/4}}{2i}$.

For $z=-i = e^{i3\pi/2}$: $\theta=3\pi/2$. $(-i)^{-1/2} = e^{-\frac{1}{2}(i3\pi/2)} = e^{-i3\pi/4}$. Residue at $z=-i$ is $\frac{e^{-i3\pi/4}}{-2i}$.

The integral along the top side of the cut is $\int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$. The integral along the bottom side is $\int_{\infty}^0 \frac{(xe^{2\pi i})^{-1/2}}{1+x^2} dx = \int_{\infty}^0 \frac{e^{-\frac{1}{2}(\ln x + 2\pi i)}}{1+x^2} dx = \int_{\infty}^0 \frac{e^{-\frac{1}{2} \ln x} e^{-\pi i}}{1+x^2} dx = \int_{\infty}^0 \frac{x^{-1/2}(-1)}{1+x^2} dx = -\int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$.

So, the sum of the integrals along the real axis parts is 0. This indicates an issue with the contour choice or branch cut definition.

Let's use the result from a reliable source for $\int_0^{\infty} \frac{x^a}{1+x^b} dx$. For $\int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$, we have $a=-1/2$ and $b=2$. The formula is $\frac{\pi}{b \sin(\frac{(a+1)\pi}{b})} = \frac{\pi}{2 \sin(\frac{(-1/2+1)\pi}{2})} = \frac{\pi}{2 \sin(\frac{\pi/2}{2})} = \frac{\pi}{2 \sin(\pi/4)} = \frac{\pi}{2 (1/\sqrt{2})} = \frac{\pi \sqrt{2}}{2}$.

So, $\int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx = \frac{\pi}{\sqrt{2}}$.

Our original integral $I$ was $2 \int_0^{\infty} \frac{x^{-1/2}}{1+x^2} dx$. Therefore,

$$I = 2 \cdot \frac{\pi}{\sqrt{2}} = \pi \sqrt{2}$$

Final Answer: The value of the integral $\int_{0}^{+\infty} \frac{\arctan x}{\sqrt{x^3}} dx$ is $\boxed{\pi\sqrt{2}}$.

This journey through series and complex analysis shows how different mathematical tools can illuminate a single problem. While the series expansion gave us a hint, it was the clever application of integration by parts followed by a contour integral (or using known results for such integrals) that ultimately unlocked the solution. Keep exploring, keep questioning, and happy integrating, guys!