Residue Theorem: Computing $\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} Dx$

Hey everyone! Today, we're diving headfirst into the fascinating world of complex analysis to tackle a pretty cool integral: $\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} dx$. This integral is a classic example of how the residue theorem can be used to calculate seemingly impossible integrals. We'll walk through the steps, break down the concepts, and hopefully, by the end, you'll feel like a total pro at this stuff. Let's get started, guys!

Diving Deep into the Integral and the Strategy

First off, let's clarify what we're dealing with. We want to find the value of $\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} dx$. This integral looks a bit intimidating at first glance, doesn't it? The presence of the absolute value $|x|$ and the complex exponential $e^{-ikx}$ suggests that we might have some fun with this one. Our main tool for today is the residue theorem. The residue theorem is a powerful tool in complex analysis that allows us to calculate the integral of a complex function along a closed contour. It tells us that the integral of a function around a closed loop is equal to $2\pi i$ times the sum of the residues of the function at the poles inside the loop. The main challenge will be setting up the right contour and figuring out how to apply the residue theorem effectively. So, let's break it down step by step.

Our primary strategy is to:

• Transform the Integral: We can start by rewriting the integral using the property that $x = re^{i\theta}$. Since we have an absolute value in the denominator, it is very important.
• Contour Integration: Construct a closed contour in the complex plane. A standard choice would be a contour that goes along the real axis and then closes in the upper or lower half-plane. We'll need to be super careful with the points where our function might blow up (poles).
• Apply the Residue Theorem: After defining the contour, we will apply the residue theorem, which says that the integral around the closed contour is equal to $2\pi i$ times the sum of the residues inside the contour. Then, we'll carefully calculate the residues at any poles that are found inside our contour.
• Evaluate the Contour Integral: The final step involves breaking the contour integral into parts and evaluating each piece separately. We'll need to think about what happens as the contour's radii go to infinity or zero. This will help us to determine which parts of the contour contribute to the final result and which ones vanish.

Setting Up the Contour and Dealing with the Branch Cut

Alright, now let's talk about setting up our contour. Because we have $|x|^{1/2}$ in the denominator, we have to deal with a branch cut. A branch cut is a line or curve in the complex plane where a multi-valued function is discontinuous. The function $z^{1/2}$ has a branch cut along the negative real axis. We will use a "keyhole contour" to avoid that branch cut. The keyhole contour typically includes a small semicircle around the origin and a large semicircle that encloses a portion of the complex plane. This contour gives us more control. The keyhole contour is crucial for handling the branch cut arising from the absolute value function. For our problem, our contour will consist of the following parts:

1. Along the Real Axis: From $-R$ to $-r$ and from $r$ to $R$, where $R$ is a large radius and $r$ is a small radius. These are the sections we are really interested in, and as $R \rightarrow \infty$ and $r \rightarrow 0$, they should give us the integral we want.
2. Large Semicircle: A large semicircle in the upper or lower half-plane with radius $R$. As $R$ tends to infinity, this part might vanish or it might not, depending on how our function behaves.
3. Small Semicircle: A small semicircle around the origin with radius $r$. As $r$ approaches zero, this part will likely be important because it will circle around the pole.
4. Around the Branch Cut: Since we are dealing with the absolute value, we need to be careful with the branch cut at the origin. The function is not well defined for negative x. The way to treat this will be to take two values from above and below the x-axis, and that's what the keyhole contour is for. The keyhole contour handles this specific discontinuity.

We need to be careful here. Remember that the absolute value $|x|$ can cause problems because the function isn't analytic at $x=0$. This is where our keyhole contour comes in handy. The crucial idea behind the keyhole contour is to go around the branch point at $x=0$. We will make a tiny loop around the origin and calculate the integral along it. The careful choice of our contour will let us isolate the parts of the integral where the function is well-behaved and where the branch cut needs special attention. Now that we've set the stage, let's get down to the nitty-gritty and calculate the actual integrals. Ready?

Applying the Residue Theorem and Evaluating the Integrals

Now for the fun part: applying the residue theorem! Let's assume, for the moment, that $k>0$. Then, we close the contour in the lower half-plane. This is because the exponential term $e^{-ikx}$ will decay in the lower half-plane, which means our integral over the large semicircle will go to zero as $R \rightarrow \infty$. If $k<0$, we would close in the upper half-plane. This ensures that the contribution from the large semicircle vanishes. The function has a branch point at $z=0$, but no poles inside our contour (besides the branch point itself). So, the residue theorem gives us:

$$\oint_C \frac{e^{-ikz}}{|z|^{1/2}} dz = 0$$

Because there are no poles inside the contour, the sum of the residues is zero. Now, we break the integral into its components, where $C$ is our keyhole contour:

$$\oint_C \frac{e^{-ikz}}{|z|^{1/2}} dz = \int_{-R}^{-r} \frac{e^{-ikx}}{(-x)^{1/2}} dx + \int_{C_r} \frac{e^{-ikz}}{|z|^{1/2}} dz + \int_{r}^{R} \frac{e^{-ikx}}{x^{1/2}} dx + \int_{C_R} \frac{e^{-ikz}}{|z|^{1/2}} dz = 0$$

Let's carefully calculate each of the integrals above:

1. Large Semicircle: As $R \rightarrow \infty$, the integral over $C_R$ vanishes. This happens because of the exponential decay in the lower half-plane, which ensures that the exponential $e^{-ikz}$ dominates and makes the integral go to zero. Specifically, the integral over the large semicircle $\int_{C_R} \frac{e^{-ikz}}{|z|^{1/2}} dz$ tends to zero as $R$ approaches infinity. The main reason here is the factor $e^{-ikz}$. Since $k>0$ and we're integrating in the lower half-plane, the exponential term causes the integral to decay rapidly to zero.
2. Small Semicircle: As $r \rightarrow 0$, the integral over $C_r$ also vanishes. This is because the singularity at the origin is of the form $1/\sqrt{z}$, which is integrable. The integral over the small semicircle $\int_{C_r} \frac{e^{-ikz}}{|z|^{1/2}} dz$ tends to zero as $r$ approaches zero. Again, we need to consider how the function behaves near the origin. In this case, the function $|z|^{1/2}$ is integrable, and so this integral goes to zero.
3. Real Axis (Parts): The key to getting the answer is to realize that the absolute value is handled by splitting the integral. As we're integrating along the negative real axis, we need to define the square root carefully using the branch cut. We write $x = -u$, with $u>0$, so $|x|^{1/2} = u^{1/2}$. This changes the integral: $\int_{-R}^{-r} \frac{e^{-ikx}}{|x|^{1/2}} dx = \int_{R}^{r} \frac{e^{iku}}{u^{1/2}} (-du) = - \int_{r}^{R} \frac{e^{iku}}{u^{1/2}} du$. On the positive real axis, we have $\int_{r}^{R} \frac{e^{-ikx}}{x^{1/2}} dx$. Now, take the limit as $r \rightarrow 0$ and $R \rightarrow \infty$, and we have: $\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} dx = \int_{0}^{\infty} \frac{e^{-ikx}}{x^{1/2}} dx - \int_{0}^{\infty} \frac{e^{ikx}}{x^{1/2}} dx$.

Therefore, we can rewrite the integral as: $\int_{0}^{\infty} \frac{e^{-ikx}}{x^{1/2}} dx + \int_{0}^{\infty} \frac{e^{ikx}}{x^{1/2}} dx = 0$. Now we apply the limits and simplify.

Final Steps and the Result

Now, let's put it all together! Our integral breaks down into two parts. After a bit of algebraic manipulation and taking the real and imaginary parts, we can find a solution. The integral on the real axis is now:

$$\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} dx = \int_{0}^{\infty} \frac{e^{-ikx}}{x^{1/2}} dx + \int_{0}^{\infty} \frac{e^{ikx}}{x^{1/2}} dx$$

Using the fact that $\int_0^\infty x^{a-1} e^{-bx} dx = b^{-a} \Gamma(a)$, and using the identity $\Gamma(1/2) = \sqrt{\pi}$ We obtain the final result:

$$\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} dx = \frac{\sqrt{2\pi}}{\sqrt{|k|}} (1-i \text{sgn}(k))$$

Where sgn(k) is the sign function.

Conclusion

So, there you have it! We've successfully used the residue theorem to compute $\int_{-\infty}^{\infty} \frac{e^{-ikx}}{|x|^{1/2}} dx$. It was a journey with a few tricky turns, but by carefully selecting our contour, handling branch cuts, and understanding the behavior of the integrals at infinity and around the origin, we got our answer. This problem shows the power of complex analysis, allowing us to solve a real integral using contour integration and the residue theorem. Keep practicing, and you'll be able to master these techniques and impress your friends. Until next time, happy integrating, guys!