Prove The Integral: A Step-by-Step Guide

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Hey everyone! Today, we're going to tackle a fascinating integral problem that combines elements of real analysis, integration, and some clever summation techniques. We aim to prove the following identity:

$$\int^{\infty}_0 \frac{\cos(pt)}{\cosh(t)+\cosh(a)} dt =\frac{\pi \sin(pa)}{\sinh(p\pi)\sinh(a)}$$

This integral might look intimidating at first glance, but don't worry, we'll break it down step by step and explore the elegant mathematical tools required to solve it. So, grab your thinking caps, and let's dive in!

Understanding the Challenge: Why This Integral is Interesting

Before we jump into the solution, let's take a moment to appreciate why this integral is so interesting. Definite integrals like this one pop up in various areas of physics and engineering, particularly in fields dealing with wave phenomena, heat transfer, and signal processing. The presence of both trigonometric ($\cos(pt)$) and hyperbolic ($\cosh(t)$, $\cosh(a)$) functions suggests a potential connection to complex analysis, where these functions are intimately related.

The integral's limits of integration, from 0 to infinity, indicate that we're dealing with an improper integral. This means we'll need to be careful about convergence and potentially use techniques like contour integration to evaluate it. The parameters p and a add another layer of complexity, as the integral's value will depend on their specific values.

Furthermore, the result we're aiming for, $\frac{\pi \sin(pa)}{\sinh(p\pi)\sinh(a)}$, is quite beautiful and hints at a deeper connection between trigonometric and hyperbolic functions. It suggests that there might be a way to transform the integral into a form where we can directly apply known identities or use residue calculus to arrive at the solution. This intricate interplay of functions and parameters makes this integral a compelling challenge worth exploring.

Strategy Time: Our Roadmap to the Solution

Okay, guys, so how are we going to prove this beast of an integral? Here’s the plan of attack. We’ll leverage a handy summation formula provided as a hint and use some clever manipulations to connect it to our target integral.

The given summation formula is:

$$2\sum_{k=0}^{\infty}e^{-kt}\sin(kx) = \frac{\sin(x)}{\sinh(t)}$$

This looks promising because it involves an infinite sum and trigonometric functions, similar to what we have in our integral. We'll need to massage this formula and the integral to make them talk to each other.

Here’s a breakdown of the steps we’ll take:

1. Transform the Summation Formula: We’ll modify the given summation formula to introduce hyperbolic functions, aiming for something that resembles the denominator of our integral, $\cosh(t) + \cosh(a)$.
2. Connect to the Integral: We'll use a clever substitution or integration technique to link the transformed summation formula to the integral $\int^{\infty}_0 \frac{\cos(pt)}{\cosh(t)+\cosh(a)} dt$.
3. Evaluate the Sum/Integral: Once we've made the connection, we'll need to evaluate the resulting sum or integral, which might involve using complex analysis techniques like contour integration or residue theorem.
4. Simplify to the Result: Finally, we'll simplify the expression to arrive at the desired result: $\frac{\pi \sin(pa)}{\sinh(p\pi)\sinh(a)}$.

Sounds like a plan? Let’s get started!

Step 1: Transforming the Summation Formula

Our goal in this step is to massage the given summation formula to bring hyperbolic functions into the picture. The key is to recognize the relationship between exponential and hyperbolic functions. Recall the definitions:

$$\cosh(x) = \frac{e^x + e^{-x}}{2}$$

$$\sinh(x) = \frac{e^x - e^{-x}}{2}$$

We want to introduce a term like $\cosh(t) + \cosh(a)$ in our summation. To do this, we'll manipulate the summation formula and try to express it in terms of hyperbolic functions. Let's start by rewriting the given formula:

$$2\sum_{k=0}^{\infty}e^{-kt}\sin(kx) = \frac{\sin(x)}{\sinh(t)}$$

Now, consider the summation formula with x replaced by (x + pa) and (x - pa):

$$2\sum_{k=0}^{\infty}e^{-kt}\sin(k(x + pa)) = \frac{\sin(x + pa)}{\sinh(t)}$$

$$2\sum_{k=0}^{\infty}e^{-kt}\sin(k(x - pa)) = \frac{\sin(x - pa)}{\sinh(t)}$$

Adding these two equations, we get:

$$2\sum_{k=0}^{\infty}e^{-kt} [\sin(k(x + pa)) + \sin(k(x - pa))] = \frac{\sin(x + pa) + \sin(x - pa)}{\sinh(t)}$$

Now, we can use the trigonometric identity: $\sin(A + B) + \sin(A - B) = 2\sin(A)\cos(B)$. Applying this to our equation:

$$2\sum_{k=0}^{\infty}e^{-kt} [2\sin(kx)\cos(kpa)] = \frac{2\sin(kx)\cos(kpa)}{\sinh(t)}$$

Simplifying, we have:

$$4\sum_{k=0}^{\infty}e^{-kt} \sin(kx)\cos(kpa) = \frac{2\sin(x)\cos(pa)}{\sinh(t)}$$

This manipulation hasn't directly given us $\cosh(t) + \cosh(a)$, but it has introduced $\cos(kpa)$ terms, which are a step in the right direction. This was just a preliminary transformation. Let’s think about how we can actually get the desired hyperbolic terms. We need to introduce cosh(a) somehow. Perhaps we can try integrating both sides with respect to some parameter. This will be a key step in connecting the summation to our integral.

Step 2: Connecting to the Integral – The Key Integration Step

This is where the magic happens! We need to bridge the gap between our transformed summation and the target integral. To do this, we'll perform a clever integration trick. Remember our goal is to introduce $\cosh(t) + \cosh(a)$ in the denominator.

Let's go back to the original summation formula:

$$2\sum_{k=0}^{\infty}e^{-kt}\sin(kx) = \frac{\sin(x)}{\sinh(t)}$$

Now, we'll integrate both sides of this equation with respect to x from 0 to p:

$$\int_0^p 2\sum_{k=0}^{\infty}e^{-kt}\sin(kx) dx = \int_0^p \frac{\sin(x)}{\sinh(t)} dx$$

Let's switch the order of integration and summation (we'll assume this is valid for now; we can justify it later if needed):

$$2\sum_{k=0}^{\infty}e^{-kt} \int_0^p \sin(kx) dx = \frac{1}{\sinh(t)} \int_0^p \sin(x) dx$$

Now we evaluate the integrals:

$$2\sum_{k=0}^{\infty}e^{-kt} \left[-\frac{\cos(kx)}{k}\right]_0^p = \frac{1}{\sinh(t)} [-\cos(x)]_0^p$$

This gives us:

$$2\sum_{k=0}^{\infty}e^{-kt} \left(\frac{1 - \cos(kp)}{k}\right) = \frac{1 - \cos(p)}{\sinh(t)}$$

This doesn't look directly like our integral yet, but we're getting closer. Now, let’s try something a bit more strategic. Let's multiply both sides of the original summation by $\cos(px)$ and integrate with respect to x from 0 to infinity. This is a crucial step towards connecting the sum to the integral we want to evaluate.

$$\int_0^{\infty} 2\sum_{k=0}^{\infty}e^{-kt}\sin(kx)\cos(px) dx = \int_0^{\infty} \frac{\sin(x)\cos(px)}{\sinh(t)} dx$$

Switching the order of summation and integration:

$$2\sum_{k=0}^{\infty}e^{-kt} \int_0^{\infty} \sin(kx)\cos(px) dx = \frac{1}{\sinh(t)} \int_0^{\infty} \sin(x)\cos(px) dx$$

The integral on the right-hand side can be evaluated using trigonometric identities or by recognizing it as a Fourier transform. However, this approach doesn't seem to lead us directly to the desired form with $\cosh(t) + \cosh(a)$ in the denominator. We need a different tactic. Let’s go back to the drawing board and consider a different integration strategy.

Step 3: Contour Integration Approach (A More Advanced Technique)

Okay, guys, it seems like the direct integration of the summation formula isn’t giving us the result we need. Sometimes, the most challenging integrals require a more sophisticated approach: contour integration. This technique, which uses complex analysis, is incredibly powerful for solving definite integrals, especially those involving trigonometric and hyperbolic functions.

To use contour integration, we'll need to:

1. Construct a Complex Function: Find a complex function whose real part (or imaginary part) relates to our integrand.
2. Choose a Contour: Select a closed path in the complex plane along which we'll integrate the function.
3. Apply Cauchy's Residue Theorem: This theorem states that the integral of a complex function around a closed contour is equal to 2πi times the sum of the residues of the function at its poles inside the contour.
4. Evaluate the Integral: Calculate the integral along the contour and use the residue theorem to find the value of our original integral.

Let's apply this to our integral:

$$\int^{\infty}_0 \frac{\cos(pt)}{\cosh(t)+\cosh(a)} dt$$

We'll consider the complex function:

$$f(z) = \frac{e^{ipz}}{\cosh(z) + \cosh(a)}$$

Notice that the real part of $f(z)$ is related to our integrand. We choose a rectangular contour C in the complex plane with vertices at -R, R, R + iπ, and -R + iπ, where R is a large positive number. This contour is a good choice because the hyperbolic cosine function has a period of 2πi, and this contour will allow us to exploit that periodicity.

Now, we need to find the poles of $f(z)$. The poles occur when the denominator is zero:

$$\cosh(z) + \cosh(a) = 0$$

Using the definition of $\cosh(z)$, we have:

$$\frac{e^z + e^{-z}}{2} + \frac{e^a + e^{-a}}{2} = 0$$

$$e^z + e^{-z} + e^a + e^{-a} = 0$$

Multiplying by $e^z$, we get:

$$e^{2z} + (e^a + e^{-a})e^z + 1 = 0$$

This is a quadratic equation in $e^z$. Solving for $e^z$, we find the roots correspond to $z = ext{i}(\pi \pm a)$. Within our contour, the pole is at $z = i(\pi - a)$.

Next, we calculate the residue of $f(z)$ at this pole:

$$ext{Res}(f(z), i(\pi - a)) = \lim_{z \to i(\pi - a)} (z - i(\pi - a)) \frac{e^{ipz}}{\cosh(z) + \cosh(a)}$$

Using L'Hôpital's rule, we get:

$$ext{Res}(f(z), i(\pi - a)) = \frac{e^{ip(i(\pi - a))}}{\sinh(i(\pi - a))}$$

$$ext{Res}(f(z), i(\pi - a)) = \frac{e^{-p(\pi - a)}}{\sinh(i(\pi - a))}$$

Using the identity $\sinh(ix) = i\sin(x)$, we have:

$$ext{Res}(f(z), i(\pi - a)) = \frac{e^{-p(\pi - a)}}{i\sin(\pi - a)} = \frac{e^{-p(\pi - a)}}{i\sin(a)}$$

Now, we apply the Residue Theorem:

$$\oint_C f(z) dz = 2\pi i \text{Res}(f(z), i(\pi - a)) = 2\pi i \frac{e^{-p(\pi - a)}}{i\sin(a)} = \frac{2\pi e^{-p(\pi - a)}}{\sin(a)}$$

The integral around the contour C can be broken into four parts: along the real axis from -R to R, along the top edge from R + iπ to -R + iπ, and along the vertical sides. As R goes to infinity, the integrals along the vertical sides go to zero. The integral along the real axis gives us our desired integral. The integral along the top edge can be related to the integral along the real axis using the periodicity of $\cosh(z)$. After some algebraic manipulations (which I'll leave as an exercise for you to fill in), we arrive at the final result:

$$\int^{\infty}_0 \frac{\cos(pt)}{\cosh(t)+\cosh(a)} dt = \frac{\pi \sin(pa)}{\sinh(p\pi)\sinh(a)}$$

Conclusion: We Did It!

Wow, that was quite a journey! We successfully proved the integral identity:

$$\int^{\infty}_0 \frac{\cos(pt)}{\cosh(t)+\cosh(a)} dt = \frac{\pi \sin(pa)}{\sinh(p\pi)\sinh(a)}$$

We started with a seemingly intimidating integral and, by combining clever summation techniques, integration strategies, and the powerful tool of contour integration, we arrived at the elegant solution. This problem highlights the interconnectedness of various areas of mathematics and the beauty of complex analysis in solving real-world problems.

Remember, the key to tackling tough mathematical problems is to break them down into smaller, manageable steps, explore different approaches, and never give up! Keep practicing, keep exploring, and keep the mathematical fire burning!