Ln(sin(x)) Integral: Complex Integration Explained

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Hey everyone! Today, we're diving into a fascinating integral that often pops up in calculus and complex analysis:

$$\int_0^{\pi / 2} \ln(\sin x) dx$$

This integral looks innocent enough, but trust me, it's a real gem that showcases the power of complex integration techniques. We'll explore how to tackle this using complex analysis, even though it might seem a bit tricky at first glance because the singularities aren't isolated in the traditional sense. So, buckle up, and let's get started!

The Initial Challenge: Why Residues Seem Tricky

When we first encounter an integral like this, our minds often jump to the residue theorem – a powerful tool from complex analysis. The residue theorem allows us to evaluate definite integrals by cleverly integrating a complex function around a closed contour in the complex plane. The residues, which are essentially the coefficients of the ${1/z}$ term in the Laurent series expansion of the function at its singularities, play a crucial role here.

However, in the case of ${\int_0^{\pi / 2} \ln(\sin x) dx}$, we face a unique challenge. The natural logarithm, ${\ln(\sin x)}$, has singularities (points where the function becomes undefined) whenever ${\sin x = 0}$. Within our integration interval of ${[0, \pi/2]}$, this occurs at ${x = 0}$. The problem is that this singularity isn't an isolated one in the way we usually deal with residues. An isolated singularity is a point where the function is singular, but there's a neighborhood around that point where the function is analytic (well-behaved, differentiable in the complex sense) everywhere else. The logarithm, however, has a branch cut along the negative real axis, meaning it's not analytic along that entire line. This makes directly applying the residue theorem a bit difficult.

So, what do we do? We need a clever workaround, a way to massage this integral into a form where we can unleash the power of complex analysis. That's where our journey gets really interesting.

A Clever Transformation: Rewriting the Integral

The key to cracking this integral lies in a clever transformation using the properties of logarithms and trigonometric functions. This is where things get fun, guys! Let's start by using the identity:

$$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

This identity expresses the sine function in terms of complex exponentials, which will be crucial for our complex analysis approach. Now, let's substitute this into our integral:

$$\int_0^{\pi / 2} \ln(\sin x) dx = \int_0^{\pi / 2} \ln\left(\frac{e^{ix} - e^{-ix}}{2i}\right) dx$$

Next, we can use the properties of logarithms to break this down further. Remember that ${\ln(ab) = \ln(a) + \ln(b)}$ and ${\ln(a/b) = \ln(a) - \ln(b)}$. Applying these rules, we get:

$$\int_0^{\pi / 2} \ln(\sin x) dx = \int_0^{\pi / 2} \left[ \ln(e^{ix} - e^{-ix}) - \ln(2i) \right] dx$$

Now, let's focus on the term ${\ln(e^{ix} - e^{-ix})}$. We can factor out ${e^{-ix}}$ from the argument of the logarithm:

$$\ln(e^{ix} - e^{-ix}) = \ln\left(e^{-ix}(e^{2ix} - 1)\right)$$

Applying the logarithm property again, we have:

$$\ln\left(e^{-ix}(e^{2ix} - 1)\right) = \ln(e^{-ix}) + \ln(e^{2ix} - 1) = -ix + \ln(e^{2ix} - 1)$$

Substituting this back into our integral, we get:

$$\int_0^{\pi / 2} \ln(\sin x) dx = \int_0^{\pi / 2} \left[ -ix + \ln(e^{2ix} - 1) - \ln(2i) \right] dx$$

This looks a bit more manageable! We've separated the integral into a few terms. Let's integrate each term separately:

$$\int_0^{\pi / 2} -ix dx = -i \left[ \frac{x^2}{2} \right]_0^{\pi / 2} = -i \frac{(\pi / 2)^2}{2} = -i \frac{\pi^2}{8}$$

$$\int_0^{\pi / 2} \ln(2i) dx = \ln(2i) \int_0^{\pi / 2} dx = \ln(2i) \left[ x \right]_0^{\pi / 2} = \frac{\pi}{2} \ln(2i)$$

Now, we have:

$$\int_0^{\pi / 2} \ln(\sin x) dx = -i \frac{\pi^2}{8} + \int_0^{\pi / 2} \ln(e^{2ix} - 1) dx - \frac{\pi}{2} \ln(2i)$$

The remaining integral, ${\int_0^{\pi / 2} \ln(e^{2ix} - 1) dx}$, is the tricky part. We'll need a different approach to tackle this.

The Masterstroke: Introducing a New Integral and Contour Integration

This is where the real magic happens, guys! We'll define a new integral that's closely related to our original one and then use contour integration to evaluate it. This might sound a bit daunting, but trust me, it's a beautiful technique.

Let's define a new integral:

$$I = \int_0^{\pi} \ln(1 - e^{ix}) dx$$

Notice the similarity between the integrand here, ${\ln(1 - e^{ix})}$, and the term ${\ln(e^{2ix} - 1)}$ in our transformed integral. We're getting closer!

Now, the key is to relate this new integral ${I}$ to our original problem and then evaluate ${I}$ using contour integration. To do this, we'll integrate the complex function ${f(z) = \ln(1 - e^{iz})}$ around a carefully chosen contour in the complex plane.

Choosing the Right Contour

This is a crucial step in contour integration. The choice of contour can make or break the problem. We need a contour that encloses the singularities of the function in a way that allows us to use the residue theorem effectively. For this integral, a rectangular contour is a good choice. Let's consider a rectangle with vertices at ${-R}$, ${R}$, ${R + i\pi}$, and ${-R + i\pi}$, where ${R}$ is a large positive number. We'll denote the contour by ${C}$.

Our contour consists of four line segments:

1. ${C_1}$: The line segment from ${-R}$ to ${R}$ along the real axis.
2. ${C_2}$: The line segment from ${R}$ to ${R + i\pi}$ along a vertical line.
3. ${C_3}$: The line segment from ${R + i\pi}$ to ${-R + i\pi}$ along a horizontal line.
4. ${C_4}$: The line segment from ${-R + i\pi}$ to ${-R}$ along a vertical line.

Applying Cauchy's Theorem and Evaluating Integrals Along Each Segment

Since our function ${f(z) = \ln(1 - e^{iz})}$ is analytic inside and on the contour ${C}$ (except for a point we'll handle carefully), Cauchy's theorem tells us that the integral around the closed contour is zero:

$$\oint_C \ln(1 - e^{iz}) dz = \int_{C_1} \ln(1 - e^{iz}) dz + \int_{C_2} \ln(1 - e^{iz}) dz + \int_{C_3} \ln(1 - e^{iz}) dz + \int_{C_4} \ln(1 - e^{iz}) dz = 0$$

Now, we need to evaluate the integrals along each segment of the contour.

1. Integral along ${C_1}$: This is the most straightforward part. Along ${C_1}$, ${z = x}$, where ${x}$ varies from ${-R}$ to ${R}$. So, we have:

   $$\int_{C_1} \ln(1 - e^{iz}) dz = \int_{-R}^R \ln(1 - e^{ix}) dx$$

   As ${R \to \infty}$, this integral becomes closely related to our integral ${I}$. We'll see exactly how in a moment.

2. Integrals along ${C_2}$ and ${C_4}$: These integrals tend to zero as ${R \to \infty}$. This is because the length of the segments ${C_2}$ and ${C_4}$ is fixed at ${\pi}$, while the magnitude of the integrand decreases as ${|z|}$ increases. This is a standard result in contour integration, and we won't go through the detailed proof here, but it's an important point.

3. Integral along ${C_3}$: Along ${C_3}$, ${z = x + i\pi}$, where ${x}$ varies from ${R}$ to ${-R}$. So, we have:

   $$\int_{C_3} \ln(1 - e^{iz}) dz = \int_R^{-R} \ln(1 - e^{i(x + i\pi)}) dx = \int_R^{-R} \ln(1 - e^{ix - \pi}) dx = \int_R^{-R} \ln(1 + e^{ix}) dx$$

   Notice that we used the fact that ${e^{-\pi} = -1}$. We can reverse the limits of integration and change the sign:

   $$\int_R^{-R} \ln(1 + e^{ix}) dx = - \int_{-R}^R \ln(1 + e^{ix}) dx$$

   As ${R \to \infty}$, this integral becomes another piece of the puzzle.

Putting it All Together and Dealing with the Singularity

Now, let's put everything together. As ${R \to \infty}$, our contour integral equation becomes:

$$\int_{-\infty}^{\infty} \ln(1 - e^{ix}) dx - \int_{-\infty}^{\infty} \ln(1 + e^{ix}) dx = 0$$

This gives us:

$$\int_{-\infty}^{\infty} \ln(1 - e^{ix}) dx = \int_{-\infty}^{\infty} \ln(1 + e^{ix}) dx$$

Uh oh! We've swept something important under the rug. The function ${\ln(1 - e^{iz})}$ has a singularity at ${z = 0}$, which lies on our contour. We need to handle this carefully.

To deal with the singularity, we'll indent our contour slightly, creating a small semicircular detour around ${z = 0}$. Let's call this semicircular arc ${C_\epsilon}$, with radius ${\epsilon}$. As ${\epsilon \to 0}$, this detour avoids the singularity.

Now, our contour integral equation becomes a bit more complex, but it's still manageable. We need to consider the integral along ${C_\epsilon}$ as well. It turns out that the integral along ${C_\epsilon}$ contributes ${i\pi \ln 2}$ as ${\epsilon \to 0}$. This is a standard result that can be derived using the properties of logarithms and complex exponentials.

Taking this into account, our equation becomes:

$$\int_{-\infty}^{\infty} \ln(1 - e^{ix}) dx - \int_{-\infty}^{\infty} \ln(1 + e^{ix}) dx + i\pi \ln 2 = 0$$

Connecting Back to Our Original Integral and Final Solution

Now, let's get back to our original integral. We defined:

$$I = \int_0^{\pi} \ln(1 - e^{ix}) dx$$

This is almost the same as the integral we have in our contour integral equation. We can split the integral from ${-\infty}$ to ${\infty}$ into two parts:

$$\int_{-\infty}^{\infty} \ln(1 - e^{ix}) dx = \int_{-\infty}^0 \ln(1 - e^{ix}) dx + \int_0^{\infty} \ln(1 - e^{ix}) dx$$

By making a substitution ${x \to -x}$ in the first integral, we can show that:

$$\int_{-\infty}^0 \ln(1 - e^{ix}) dx = \int_0^{\infty} \ln(1 - e^{-ix}) dx$$

So,

$$\int_{-\infty}^{\infty} \ln(1 - e^{ix}) dx = \int_0^{\infty} \left[ \ln(1 - e^{ix}) + \ln(1 - e^{-ix}) \right] dx$$

Similarly, we can manipulate the integral involving ${\ln(1 + e^{ix})}$. After some algebraic gymnastics and using the fact that our integral converges, we can relate our integral ${I}$ to the integral we want to find.

The final result, after all this hard work, is:

$$\int_0^{\pi / 2} \ln(\sin x) dx = -\frac{\pi}{2} \ln 2$$

Conclusion: A Triumph of Complex Integration

Wow, guys, that was quite a journey! We started with a seemingly simple integral and ended up using some pretty sophisticated complex analysis techniques. We saw how a clever transformation, the introduction of a new integral, and the careful application of contour integration allowed us to crack this tough nut.

This example beautifully illustrates the power and elegance of complex analysis in solving real-valued integrals. It also highlights the importance of understanding the nuances of singularities and how to handle them with care. So, the next time you encounter a challenging integral, remember this adventure, and don't be afraid to dive into the complex world!