Contour Integration: Evaluating Tricky Integrals

by Axel Sørensen 49 views

Hey guys! Let's dive into the fascinating world of contour integration and how we can use it to solve some tricky integrals. Today, we're tackling the integral:

0arctan(η2+x)(x+η1)2+η32dx\displaystyle\int\limits_{0}^\infty\frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

where η1{\eta_1}, η2{\eta_2}, and η3{\eta_3} are real numbers. This integral looks a bit intimidating, but with the right contour, we can tame it!

Understanding the Integral and the Challenge

Before we jump into contour integration, let's break down what we're dealing with. The integral involves an arctangent function and a rational function, making it a bit challenging to solve using elementary techniques. This is where contour integration comes to the rescue.

Contour integration is a powerful tool that allows us to evaluate real integrals by extending them into the complex plane. The basic idea is to find a suitable contour (a closed path in the complex plane), integrate the complex function along this contour, and then use the residue theorem to relate the contour integral to the original real integral. So, the million-dollar question is: what kind of contour should we use for this particular integral?

Choosing the Right Contour: The Key to Success

Selecting the right contour is crucial for successfully evaluating an integral using contour integration. The contour should be chosen such that:

  1. The integral along the contour can be easily evaluated.
  2. The contour encloses singularities (poles) of the integrand.
  3. The original real integral is a part of the contour integral.

For integrals of this form, a semi-circular contour in the upper half-plane is often a good choice. Let's explore why this might work and what considerations we need to keep in mind.

Why a Semi-Circular Contour?

A semi-circular contour, denoted by C{C}, typically consists of two parts:

  • A straight line segment, C1{C_1}, along the real axis from R{-R} to R{R}, where R{R} is a large positive number.
  • A semi-circular arc, C2{C_2}, in the upper half-plane, centered at the origin, with radius R{R}.

This contour is appealing because:

  • The integral along the real axis segment C1{C_1} will be related to our original integral as R{R} approaches infinity.
  • The semi-circular arc C2{C_2} allows us to use the residue theorem, provided we can evaluate the integral along this arc.

Considerations for Our Specific Integral

For our specific integral, we need to consider the following:

  • Singularities: We need to identify the singularities of the integrand in the complex plane. The denominator (z+η1)2+η32{(z+\eta_1)^2 + \eta_3^2} gives us potential singularities. Let's find them by setting the denominator to zero: (z+η1)2+η32=0{(z+\eta_1)^2 + \eta_3^2 = 0} (z+η1)2=η32{(z+\eta_1)^2 = -\eta_3^2} z+η1=±iη3{z+\eta_1 = \pm i\eta_3} z=η1±iη3{z = -\eta_1 \pm i\eta_3} So, we have two singularities: z1=η1+iη3{z_1 = -\eta_1 + i\eta_3} and z2=η1iη3{z_2 = -\eta_1 - i\eta_3}. Since we're considering a semi-circular contour in the upper half-plane, we only need to worry about singularities with a positive imaginary part. Thus, we focus on z1=η1+iη3{z_1 = -\eta_1 + i\eta_3}. For the contour to enclose the singularity, we need η3>0{\eta_3 > 0}.
  • Behavior of the Integrand: We need to ensure that the integral along the semi-circular arc C2{C_2} vanishes as R{R} approaches infinity. This often requires bounding the integrand and showing that it decays sufficiently rapidly as z{|z|} becomes large. For our integral, we need to analyze the behavior of arctan(η2+z)(z+η1)2+η32{\frac{\arctan(\eta_2+z)}{(z+\eta_1)^2+\eta_3^2}} as z{|z| \to \infty} in the upper half-plane. The arctangent function is bounded, and the denominator grows quadratically, which is promising.
  • Branch Cuts: The arctangent function is multi-valued in the complex plane and requires a branch cut. The standard branch cuts for arctan(z){\arctan(z)} are along the imaginary axis from i{i} to i{i\infty} and from i{-i} to i{-i\infty}. We need to make sure that our contour doesn't intersect these branch cuts, which means carefully choosing the parameters η2{\eta_2} and R{R}.

Setting Up the Contour Integral

Now that we've discussed the considerations, let's set up the contour integral. We'll integrate the complex function

f(z)=arctan(η2+z)(z+η1)2+η32{f(z) = \frac{\arctan(\eta_2+z)}{(z+\eta_1)^2+\eta_3^2}}

around the semi-circular contour C{C}. According to the residue theorem:

Cf(z)dz=2πiRes(f,zk)\oint_C f(z)\, dz = 2\pi i \sum \text{Res}(f, z_k)

where the sum is over the residues of f{f} at the singularities zk{z_k} enclosed by the contour C{C}. In our case, assuming η3>0{\eta_3 > 0} and that z1=η1+iη3{z_1 = -\eta_1 + i\eta_3} is inside the contour, we have:

Cf(z)dz=2πiRes(f,z1)\oint_C f(z)\, dz = 2\pi i \text{Res}(f, z_1)

We can write the contour integral as the sum of the integrals along the straight line segment C1{C_1} and the semi-circular arc C2{C_2}:

Cf(z)dz=C1f(z)dz+C2f(z)dz\oint_C f(z)\, dz = \int_{C_1} f(z)\, dz + \int_{C_2} f(z)\, dz

Our goal is to relate the integral along C1{C_1} to the original real integral and show that the integral along C2{C_2} vanishes as R{R \to \infty}.

Evaluating the Integrals Along C1{C_1} and C2{C_2}

Integral Along C1{C_1}

Along C1{C_1}, we have z=x{z = x}, where x{x} varies from R{-R} to R{R}. So,

C1f(z)dz=RRarctan(η2+x)(x+η1)2+η32dx\int_{C_1} f(z)\, dz = \int_{-R}^R \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

As R{R \to \infty}, this integral approaches:

limRRRarctan(η2+x)(x+η1)2+η32dx\lim_{R \to \infty} \int_{-R}^R \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

This is closely related to our original integral, but it's an integral from {-\infty} to {\infty}, while our original integral is from 0{0} to {\infty}. We'll need to address this difference later.

Integral Along C2{C_2}

Along C2{C_2}, we parameterize z{z} as z=Reiθ{z = Re^{i\theta}}, where 0θπ{0 \leq \theta \leq \pi}. Then, dz=iReiθdθ{dz = iRe^{i\theta}\, d\theta}, and the integral becomes:

C2f(z)dz=0πarctan(η2+Reiθ)(Reiθ+η1)2+η32iReiθdθ\int_{C_2} f(z)\, dz = \int_{0}^\pi \frac{\arctan(\eta_2+Re^{i\theta})}{(Re^{i\theta}+\eta_1)^2+\eta_3^2} iRe^{i\theta}\, d\theta

To show that this integral vanishes as R{R \to \infty}, we need to bound the integrand. We know that arctan(z){|\arctan(z)|} is bounded in the upper half-plane, and the denominator grows like R2{R^2}. Therefore, the integrand behaves like 1R2{\frac{1}{R^2}}, and the integral behaves like RR2=1R{\frac{R}{R^2} = \frac{1}{R}}, which approaches 0 as R{R \to \infty}. This is a crucial step in ensuring that our contour integration method works.

Calculating the Residue

Now, let's calculate the residue of f(z){f(z)} at the singularity z1=η1+iη3{z_1 = -\eta_1 + i\eta_3}. Since z1{z_1} is a simple pole (a pole of order 1), the residue is given by:

Res(f,z1)=limzz1(zz1)f(z)\text{Res}(f, z_1) = \lim_{z \to z_1} (z - z_1) f(z)

Res(f,z1)=limzη1+iη3(z(η1+iη3))arctan(η2+z)(z+η1)2+η32\text{Res}(f, z_1) = \lim_{z \to -\eta_1 + i\eta_3} (z - (- \eta_1 + i\eta_3)) \frac{\arctan(\eta_2+z)}{(z+\eta_1)^2+\eta_3^2}

We can rewrite the denominator as:

(z+η1)2+η32=(z+η1+iη3)(z+η1iη3){(z+\eta_1)^2+\eta_3^2 = (z+\eta_1 + i\eta_3)(z+\eta_1 - i\eta_3)}

So,

Res(f,z1)=limzη1+iη3(z+η1iη3)arctan(η2+z)(z+η1+iη3)(z+η1iη3)\text{Res}(f, z_1) = \lim_{z \to -\eta_1 + i\eta_3} (z + \eta_1 - i\eta_3) \frac{\arctan(\eta_2+z)}{(z+\eta_1 + i\eta_3)(z+\eta_1 - i\eta_3)}

Res(f,z1)=limzη1+iη3arctan(η2+z)z+η1+iη3\text{Res}(f, z_1) = \lim_{z \to -\eta_1 + i\eta_3} \frac{\arctan(\eta_2+z)}{z+\eta_1 + i\eta_3}

Plugging in z=η1+iη3{z = -\eta_1 + i\eta_3}, we get:

Res(f,z1)=arctan(η2η1+iη3)η1+iη3+η1+iη3=arctan(η2η1+iη3)2iη3\text{Res}(f, z_1) = \frac{\arctan(\eta_2 - \eta_1 + i\eta_3)}{-\eta_1 + i\eta_3 + \eta_1 + i\eta_3} = \frac{\arctan(\eta_2 - \eta_1 + i\eta_3)}{2i\eta_3}

Putting It All Together

Now we have all the pieces. From the residue theorem:

Cf(z)dz=2πiRes(f,z1)=2πiarctan(η2η1+iη3)2iη3=πη3arctan(η2η1+iη3)\oint_C f(z)\, dz = 2\pi i \text{Res}(f, z_1) = 2\pi i \frac{\arctan(\eta_2 - \eta_1 + i\eta_3)}{2i\eta_3} = \frac{\pi}{\eta_3} \arctan(\eta_2 - \eta_1 + i\eta_3)

And from the contour integral decomposition:

Cf(z)dz=limRRRarctan(η2+x)(x+η1)2+η32dx+limRC2f(z)dz\oint_C f(z)\, dz = \lim_{R \to \infty} \int_{-R}^R \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx + \lim_{R \to \infty} \int_{C_2} f(z)\, dz

Since the integral along C2{C_2} vanishes as R{R \to \infty}, we have:

πη3arctan(η2η1+iη3)=arctan(η2+x)(x+η1)2+η32dx\frac{\pi}{\eta_3} \arctan(\eta_2 - \eta_1 + i\eta_3) = \int_{-\infty}^\infty \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

Addressing the Integration Limits

We've evaluated the integral from {-\infty} to {\infty}, but our original integral is from 0{0} to {\infty}. To relate these, we can split the integral:

arctan(η2+x)(x+η1)2+η32dx=0arctan(η2+x)(x+η1)2+η32dx+0arctan(η2+x)(x+η1)2+η32dx\int_{-\infty}^\infty \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx = \int_{-\infty}^0 \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx + \int_0^\infty \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

Now, we can make a substitution x=u{x = -u} in the first integral:

0arctan(η2+x)(x+η1)2+η32dx=0arctan(η2u)(u+η1)2+η32(du)=0arctan(η2x)(xη1)2+η32dx\int_{-\infty}^0 \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx = \int_{\infty}^0 \frac{\arctan(\eta_2-u)}{(-u+\eta_1)^2+\eta_3^2} (-du) = \int_0^\infty \frac{\arctan(\eta_2-x)}{(x-\eta_1)^2+\eta_3^2}\, dx

So, we have:

πη3arctan(η2η1+iη3)=0arctan(η2x)(xη1)2+η32dx+0arctan(η2+x)(x+η1)2+η32dx\frac{\pi}{\eta_3} \arctan(\eta_2 - \eta_1 + i\eta_3) = \int_0^\infty \frac{\arctan(\eta_2-x)}{(x-\eta_1)^2+\eta_3^2}\, dx + \int_0^\infty \frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

Unfortunately, this doesn't directly give us our original integral. The integral we want is:

0arctan(η2+x)(x+η1)2+η32dx\int_0^\infty\frac{\arctan(\eta_2+x)}{(x+\eta_1)^2+\eta_3^2}\, dx

To get this, we would need to isolate this term, which requires further manipulation and analysis of the resulting integrals. This might involve looking for symmetries or additional relationships between the integrals.

Final Thoughts and Next Steps

We've made significant progress in evaluating the integral using contour integration. We've identified the appropriate contour (a semi-circle in the upper half-plane), calculated the residue, and related the contour integral to the original integral. However, we've also encountered a challenge in relating the integral from {-\infty} to {\infty} to our desired integral from 0{0} to {\infty}.

To fully solve this problem, we might need to:

  1. Explore additional techniques for simplifying the integral expression.
  2. Consider different contours or variations of the semi-circular contour.
  3. Analyze the properties of the arctangent function more deeply.

Contour integration can be a complex but incredibly powerful tool, so keep practicing, guys! This problem illustrates the key steps and considerations involved in using this method. Keep pushing, and you'll master these techniques in no time!