Evaluating The Integral Of (sqrt(x) / (x^2 + 1)) * Log((x + 1) / (2 * Sqrt(x)))

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Hey everyone! Today, we're going to tackle a fascinating integral problem that combines the elegance of complex analysis with the power of definite integrals. We'll be diving deep into the evaluation of $\int_0^\infty \frac\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right);dx$, and trust me, it's going to be a rewarding journey. Our ultimate goal is to prove that this integral equals a rather intriguing expression $\frac{\pi\sqrt{2}{2}\log\left(1+\frac{\sqrt{2}}{2}\right)$. Buckle up, because we're about to embark on a mathematical adventure!

The Challenge: A Tricky Integral

When we first look at this integral, it might seem a bit daunting. We have a square root, a rational function, and a logarithm all mixed together. It's not your typical straightforward integration problem. The presence of the logarithm, especially, hints that we might need some clever techniques to crack this nut. We can't just jump in with standard calculus methods; we need a strategic approach. Complex analysis, with its powerful tools like contour integration, often comes to the rescue in such scenarios. The logarithmic term and the form of the integrand strongly suggest that contour integration might be a viable path forward.

The integral's limits, from 0 to infinity, are another clue. This type of integral often lends itself well to contour integration techniques, where we close the integration path in the complex plane. But before we get into the complex stuff, let's take a moment to appreciate the beauty of this problem. It's a testament to how different areas of mathematics can come together to solve a single, elegant question. This integral isn't just about finding a numerical answer; it's about the journey, the methods, and the insights we gain along the way. So, let's roll up our sleeves and get started!

Setting the Stage: Complex Analysis and Contour Integration

Complex analysis is our primary weapon in this battle against the integral. For those who aren't as familiar, complex analysis is the branch of mathematics that deals with functions of complex numbers. It provides a powerful toolkit for solving problems that are difficult or impossible to solve using real analysis alone. One of the most potent tools in this kit is contour integration. Contour integration involves integrating a complex function along a path (or contour) in the complex plane. The beauty of this technique lies in its ability to transform real integrals into complex ones, which can often be evaluated more easily using the residue theorem or other complex analysis techniques.

Now, why is contour integration so useful here? Well, the integrand $\frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)$ behaves nicely in the complex plane, except at a few specific points (singularities). We can carefully choose a contour that avoids these singularities and then use the residue theorem to relate the integral along the contour to the residues of the function at the enclosed singularities. The residue theorem, in essence, says that the integral of a complex function around a closed curve is proportional to the sum of the residues of the function at the singularities enclosed by the curve. This theorem is the heart of contour integration, allowing us to turn a complex integral into a simple algebraic calculation.

Choosing the right contour is crucial. For integrals over the real line, a common choice is a semi-circular contour in the upper or lower half-plane. However, due to the presence of the square root and logarithm, we might need a more creative contour, such as a keyhole contour, which wraps around the branch cut of the logarithm. The keyhole contour is especially useful when dealing with multi-valued functions like the square root and logarithm, as it allows us to carefully track the changes in the function's value as we circle the origin. Before diving into the specifics of our chosen contour, let's recap the plan: we'll transform our real integral into a complex one, choose a suitable contour, identify the singularities, calculate the residues, and then use the residue theorem to evaluate the integral. Sounds like a plan, right?

The Keyhole Contour: Our Secret Weapon

Alright, let's talk about our secret weapon: the keyhole contour. This is a clever choice when dealing with integrals involving fractional powers (like our square root) and logarithms. The keyhole contour, as the name suggests, resembles a keyhole shape in the complex plane. It consists of two circular arcs, one large (radius R) and one small (radius r), connected by two line segments running parallel to the real axis. The small circle is centered at the origin, which is often a branch point for functions like $\sqrt{z}$ and $\log(z)$. The two line segments lie just above and below the positive real axis, creating a "keyhole" that allows us to navigate the branch cut of the logarithm.

Why is this contour so effective? The main reason is that it allows us to handle the multi-valued nature of the complex logarithm and the square root. As we traverse the contour, the argument of the complex variable changes, and the logarithm picks up multiples of $2\pi i$. The keyhole contour cleverly uses this fact to our advantage. By integrating along both sides of the branch cut, we can relate the integrals in a way that cancels out some of the unwanted terms and isolates the integral we're interested in. Imagine walking along the contour: as you go around the origin, the logarithm changes its value, but the keyhole shape ensures that we can keep track of these changes and use them to our benefit.

Specifically, our keyhole contour, denoted by C, will consist of the following parts:

1. $C_R$: A circular arc with radius R in the upper half-plane, going counterclockwise from angle $\epsilon$ to $2\pi - \epsilon$.
2. $C_r$: A circular arc with radius r in the upper half-plane, going clockwise from angle $2\pi - \epsilon$ to $\epsilon$.
3. $L_1$: A line segment along the positive real axis from r to R.
4. $L_2$: A line segment along the positive real axis from R to r, but just above the real axis.

As we let R go to infinity and r go to zero, the integrals along $C_R$ and $C_r$ will vanish (we'll prove this later), leaving us with the integrals along $L_1$ and $L_2$. These integrals will be related to our original integral, and we'll be able to solve for it using the residue theorem. The keyhole contour might seem a bit intricate, but it's a powerful tool in our arsenal, allowing us to conquer this challenging integral. So, let's move on to the next step: defining our complex function and identifying its singularities.

Defining the Complex Function and Identifying Singularities

Now that we have our keyhole contour ready, let's define the complex function we'll be integrating. This is a crucial step because the choice of function directly impacts how we apply the residue theorem. We need a function that captures the essence of our original integral while being amenable to complex analysis techniques. A natural choice is to replace the real variable 'x' with the complex variable 'z' and consider the function:

$$f(z) = \frac{\sqrt{z}}{z^2+1}\log\left(\frac{z+1}{2\sqrt{z}}\right)$$

This function looks very similar to our original integrand, but now it's defined for complex numbers. Notice that we have a square root and a logarithm, both of which are multi-valued functions in the complex plane. This is why we chose the keyhole contour – to carefully handle these multi-valuedness issues. The square root function, $\sqrt{z}$, has a branch point at $z=0$, and the logarithm, $\log\left(\frac{z+1}{2\sqrt{z}}\right)$, also has a branch point at $z=0$ due to the $\sqrt{z}$ term in its argument. Additionally, the logarithm has a branch point where its argument is zero or infinity. Let's analyze the argument:

$$\frac{z+1}{2\sqrt{z}}$$

This argument is zero when $z+1=0$, which means $z=-1$. It's infinity when $\sqrt{z}=0$, which we already know is $z=0$. So, we have branch points at $z=0$ and $z=-1$. However, since our keyhole contour already encircles the branch point at $z=0$, we need to be mindful of the branch point at $z=-1$ as well. But for the purpose of applying the residue theorem, we primarily focus on the poles of the function. The poles occur where the denominator is zero, which is $z^2+1=0$. This gives us $z = \pm i$, which are simple poles located at $i$ and $-i$ in the complex plane.

These poles are crucial because the residue theorem tells us that the integral around a closed contour is related to the residues at the poles enclosed by the contour. Our keyhole contour encloses the pole at $z=i$, but not at $z=-i$. Therefore, we only need to calculate the residue at $z=i$. Remember, the residue of a function at a pole is a measure of the function's singularity at that point. It's a key ingredient in the residue theorem, allowing us to connect the complex integral to a simple algebraic quantity. So, next up, we'll calculate the residue of our function at the pole $z=i$.

Calculating the Residue at z = i

Okay, time to get our hands dirty and calculate the residue of our complex function $f(z) = \frac{\sqrt{z}}{z^2+1}\log\left(\frac{z+1}{2\sqrt{z}}\right)$ at the pole $z = i$. Remember, the residue is a crucial value that helps us connect the complex integral to the singularities of the function. Since $z=i$ is a simple pole (meaning the denominator has a factor of $(z-i)$ but not $(z-i)^2$ or higher powers), we can use a straightforward formula to calculate the residue. For a simple pole at $z=z_0$, the residue is given by:

$$Res(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)$$

In our case, $z_0 = i$, so we have:

$$Res(f, i) = \lim_{z \to i} (z - i)\frac{\sqrt{z}}{z^2+1}\log\left(\frac{z+1}{2\sqrt{z}}\right)$$

We can rewrite the denominator as $z^2 + 1 = (z - i)(z + i)$, which simplifies our expression:

$$Res(f, i) = \lim_{z \to i} (z - i)\frac{\sqrt{z}}{(z - i)(z + i)}\log\left(\frac{z+1}{2\sqrt{z}}\right) = \lim_{z \to i} \frac{\sqrt{z}}{z + i}\log\left(\frac{z+1}{2\sqrt{z}}\right)$$

Now, we can directly substitute $z = i$ into the expression:

$$Res(f, i) = \frac{\sqrt{i}}{i + i}\log\left(\frac{i+1}{2\sqrt{i}}\right) = \frac{\sqrt{i}}{2i}\log\left(\frac{i+1}{2\sqrt{i}}\right)$$

To simplify this further, we need to express $\sqrt{i}$ and $\frac{i+1}{2\sqrt{i}}$ in polar form. Remember that $i = e^{i\pi/2}$, so $\sqrt{i} = e^{i\pi/4} = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$. Also, $1 + i = \sqrt{2}e^{i\pi/4}$. Therefore:

$$\frac{i+1}{2\sqrt{i}} = \frac{\sqrt{2}e^{i\pi/4}}{2e^{i\pi/4}} = \frac{\sqrt{2}}{2}$$

Plugging these back into our residue expression, we get:

$$Res(f, i) = \frac{\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}}{2i}\log\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}(1 + i)}{4i}\log\left(\frac{\sqrt{2}}{2}\right)$$

This is a complex number, and we'll use it in the residue theorem to relate the integral around our keyhole contour to the value we're trying to find. We've successfully navigated the tricky calculation of the residue at $z=i$, which is a major step forward in our quest to evaluate the integral.

Applying the Residue Theorem and Evaluating Contour Integrals

With the residue at $z = i$ in hand, we're now ready to unleash the Residue Theorem. This powerful theorem is the linchpin of our approach, connecting the integral around the closed keyhole contour C to the sum of the residues of our function f(z) at the poles enclosed by C. The Residue Theorem states:

$$\oint_C f(z) dz = 2\pi i \sum Res(f, z_k)$$

where the sum is taken over all poles $z_k$ of $f(z)$ that lie inside the contour C. In our case, the keyhole contour encloses only one pole, $z = i$. So, we have:

$$\oint_C f(z) dz = 2\pi i \cdot Res(f, i)$$

We already calculated the residue at $z = i$ to be:

$$Res(f, i) = \frac{\sqrt{2}(1 + i)}{4i}\log\left(\frac{\sqrt{2}}{2}\right)$$

Therefore:

$$\oint_C f(z) dz = 2\pi i \cdot \frac{\sqrt{2}(1 + i)}{4i}\log\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi\sqrt{2}(1 + i)}{2}\log\left(\frac{\sqrt{2}}{2}\right)$$

Now, remember that our contour C consists of four parts: $C_R$, $C_r$, $L_1$, and $L_2$. So, we can write the contour integral as a sum of integrals over these parts:

$$\oint_C f(z) dz = \int_{C_R} f(z) dz + \int_{C_r} f(z) dz + \int_{L_1} f(z) dz + \int_{L_2} f(z) dz$$

We'll now evaluate the integrals along each part of the contour. First, we'll show that the integrals along the large circle $C_R$ and the small circle $C_r$ vanish as their radii go to infinity and zero, respectively. This will leave us with the integrals along the line segments $L_1$ and $L_2$, which are directly related to our original integral.

Vanishing Integrals on $C_R$ and $C_r$

Let's first show that $\int_{C_R} f(z) dz$ vanishes as $R \to \infty$. On $C_R$, we have $z = Re^{i\theta}$, where $\theta$ varies from $\epsilon$ to $2\pi - \epsilon$. As $R \to \infty$, we have:

$$|f(z)| = \left|\frac{\sqrt{z}}{z^2+1}\log\left(\frac{z+1}{2\sqrt{z}}\right)\right| \approx \frac{\sqrt{R}}{R^2} \left| \log\left(\frac{Re^{i\theta}}{2\sqrt{R}e^{i\theta/2}}\right) \right| \approx \frac{1}{R^{3/2}} |\log(R)|$$

Since $\lim_{R \to \infty} \frac{\log(R)}{R^{3/2}} = 0$, the integral $\int_{C_R} f(z) dz$ vanishes as $R \to \infty$ by the estimation lemma.

Next, we'll show that $\int_{C_r} f(z) dz$ vanishes as $r \to 0$. On $C_r$, we have $z = re^{i\theta}$, where $\theta$ varies from $2\pi - \epsilon$ to $\epsilon$. As $r \to 0$, we have:

$$|f(z)| = \left|\frac{\sqrt{z}}{z^2+1}\log\left(\frac{z+1}{2\sqrt{z}}\right)\right| \approx \frac{\sqrt{r}}{1} |\log(r)|$$

Since $\lim_{r \to 0} \sqrt{r}\log(r) = 0$, the integral $\int_{C_r} f(z) dz$ vanishes as $r \to 0$.

Integrals on $L_1$ and $L_2$

Now we're left with the integrals along $L_1$ and $L_2$. On $L_1$, we have $z = x$, where $x$ varies from $r$ to $R$. Thus:

$$\int_{L_1} f(z) dz = \int_r^R \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx$$

On $L_2$, we also have $z = x$, but since we're just above the branch cut, we need to account for the change in the argument of the square root. We have $\sqrt{z} = \sqrt{x}e^{i\pi}$, and the logarithm picks up a $2\pi i$ term. So:

$$\int_{L_2} f(z) dz = -\int_r^R \frac{\sqrt{x}e^{i\pi}}{x^2+1}\left[\log\left(\frac{x+1}{2\sqrt{x}e^{i\pi}}\right)\right] dx = \int_R^r \frac{-\sqrt{x}}{x^2+1}\left[\log\left(\frac{x+1}{2\sqrt{x}}e^{-i\pi}\right)\right] dx$$

Combining these results with the Residue Theorem, we can finally solve for our integral.

Final Calculation and the Grand Finale

We've reached the final act of our mathematical play! We've set the stage with complex analysis, chosen our keyhole contour, calculated the residue, and applied the residue theorem. Now, it's time to put all the pieces together and arrive at our grand finale: the value of the integral. Remember, we have:

$$\oint_C f(z) dz = \int_{C_R} f(z) dz + \int_{C_r} f(z) dz + \int_{L_1} f(z) dz + \int_{L_2} f(z) dz$$

We've shown that the integrals along $C_R$ and $C_r$ vanish as $R \to \infty$ and $r \to 0$. So, we're left with:

$$\oint_C f(z) dz = \int_{L_1} f(z) dz + \int_{L_2} f(z) dz$$

We also know from the residue theorem that:

$$\oint_C f(z) dz = \frac{\pi\sqrt{2}(1 + i)}{2}\log\left(\frac{\sqrt{2}}{2}\right)$$

Now, let's look at the integrals along $L_1$ and $L_2$. As we found earlier:

$$\int_{L_1} f(z) dz = \int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx$$

$$\int_{L_2} f(z) dz = \int_\infty^0 \frac{-\sqrt{x}}{x^2+1}\left[\log\left(\frac{x+1}{2\sqrt{x}}e^{-i\pi}\right)\right] dx = \int_0^\infty \frac{\sqrt{x}}{x^2+1}\left[\log\left(\frac{x+1}{2\sqrt{x}}\right) - i\pi\right] dx$$

Adding these two integrals, we get:

$$\oint_C f(z) dz = 2\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx - i\pi \int_0^\infty \frac{\sqrt{x}}{x^2+1} dx$$

Now, we can equate the real and imaginary parts of this expression with the result from the residue theorem:

$$\frac{\pi\sqrt{2}(1 + i)}{2}\log\left(\frac{\sqrt{2}}{2}\right) = 2\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx - i\pi \int_0^\infty \frac{\sqrt{x}}{x^2+1} dx$$

Equating the real parts, we have:

$$\frac{\pi\sqrt{2}}{2}\log\left(\frac{\sqrt{2}}{2}\right) = 2\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx$$

Therefore:

$$\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx = \frac{\pi\sqrt{2}}{4}\log\left(\frac{\sqrt{2}}{2}\right)$$

But wait! We're not quite there yet. We need to use the fact that $\log(\frac{\sqrt{2}}{2}) = \log(\frac{1}{\sqrt{2}}) = -\log(\sqrt{2}) = -\frac{1}{2}\log(2)$. Also, we need to remember our target result: $\frac{\pi\sqrt{2}}{2}\log\left(1+\frac{\sqrt{2}}{2}\right)$. We made an error in the residue calculation or somewhere else. Let's try another approach. Equating imaginary parts:

\frac{\pi\sqrt{2}}{2}\log\left(\frac{\sqrt{2}}{2}\right) = -\pi \int_0^\infty \frac{\sqrt{x}}{x^2+1} dx$$\int_0^\infty \frac{\sqrt{x}}{x^2+1} dx = -\frac{\sqrt{2}}{2}\log\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}\log(\sqrt{2}) = \frac{\sqrt{2}}{4}\log(2)

This is a known integral. Let's go back to the real part equation:

$$\frac{\pi\sqrt{2}}{4}\log\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi\sqrt{2}}{4}(- \log \sqrt{2}) = -\frac{\pi\sqrt{2}}{8} \log 2$$

We need to find the correct expression. We had:

$$2\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx = Re\left[2\pi i \frac{\sqrt{i}}{2i} \log\left(\frac{1+i}{2\sqrt{i}}\right)\right]$$

$$\sqrt{i} = e^{i\pi/4} = \frac{1+i}{\sqrt{2}}$$

$$\frac{1+i}{2\sqrt{i}} = \frac{1+i}{2(1+i)/\sqrt{2}} = \frac{\sqrt{2}}{2}$$

$$\log(\sqrt{2}/2) = \log(2^{-1/2}) = -\frac{1}{2}\log 2$$

$$Res(f, i) = \frac{\sqrt{i}}{2i} \log(\frac{1+i}{2\sqrt{i}}) = \frac{(1+i)/\sqrt{2}}{2i} (- \frac{1}{2} \log 2) = \frac{1+i}{2i\sqrt{2}} (- \frac{1}{2} \log 2) = \frac{(1+i)(-i)}{2\sqrt{2}} (- \frac{1}{2} \log 2) = \frac{1-i}{2\sqrt{2}} (- \frac{1}{2} \log 2)$$

$$2 \pi i Res = 2 \pi i \frac{1-i}{2\sqrt{2}} (- \frac{1}{2} \log 2) = \frac{\pi i (1-i)}{\sqrt{2}} (- \frac{1}{2} \log 2) = \frac{\pi (1+i)}{2\sqrt{2}} \log 2$$

$$Re[2 \pi i Res] = \frac{\pi \log 2}{2\sqrt{2}}$$

$$\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) dx = \frac{\pi \log 2}{4\sqrt{2}}$$

Oops! Still incorrect...I acknowledge that there might be subtle errors in the above calculations. Arriving at the desired final result $\frac{\pi\sqrt{2}}{2}\log\left(1+\frac{\sqrt{2}}{2}\right)$ requires careful consideration of branches, the correct residues and algebraic manipulations. While the outline of using the keyhole contour and the residue theorem is correct, the final steps need to be meticulously checked. If I were to continue, I would revisit the residue calculation, double-check the argument of the logarithm, and carefully consider the behavior of each term as $z$ approaches $i$.

Conclusion: A Challenging but Rewarding Journey

Even though we haven't quite reached the final answer in this attempt, we've embarked on a fascinating journey through the world of complex analysis and definite integrals. We've seen how the keyhole contour and the residue theorem can be powerful tools for tackling tricky integrals. While there might be some errors in the algebraic details of our calculation, the overall strategy is sound. This problem serves as a great reminder that mathematics is not just about getting the right answer; it's about the process, the techniques, and the insights we gain along the way. So, keep exploring, keep questioning, and keep solving!