can you find the residue at s=-(((2*n - 1)*Pi)/(2*l))^2 ? Please provide the step-by-step solution.

sinh(sqrt(s)*x)*exp(s*t)/(s*cosh(sqrt(s)*l)*sqrt(s))

Certainly! Let's compute the residue of the given function at the point \( s = -\left(\frac{(2n - 1)\pi}{2l}\right)^2 \).

### Step 1: Identify the function and the point

The given function is:

\[
f(s) = \frac{\sinh(\sqrt{s} \cdot x) \cdot \exp(s \cdot t)}{s \cosh(\sqrt{s} \cdot l) \cdot \sqrt{s}}
\]

We are interested in finding the residue at the point \( s_0 = -\left(\frac{(2n - 1)\pi}{2l}\right)^2 \).

### Step 2: Identify the nature of the singularity

First, let's rewrite \( s_0 \) for clarity:

\[
s_0 = -\left(\frac{(2n - 1)\pi}{2l}\right)^2
\]

Substituting \( s_0 \) into the square root:

\[
\sqrt{s_0} = i \cdot \frac{(2n - 1)\pi}{2l}
\]

### Step 3: Simplify the function around the pole

We expand the function \( f(s) \) around \( s = s_0 \):

\[
f(s) = \frac{\sinh(\sqrt{s} \cdot x) \cdot \exp(s \cdot t)}{s \cosh(\sqrt{s} \cdot l) \cdot \sqrt{s}}
\]

Substitute \( \sqrt{s} = \sqrt{s_0 + \delta s} = \sqrt{s_0} \sqrt{1 + \frac{\delta s}{s_0}} \) where \( \delta s = s - s_0 \).

For small \( \delta s \), we can use the approximation:

\[
\sqrt{1 + \frac{\delta s}{s_0}} \approx 1 + \frac{\delta s}{2s_0}
\]

This gives:

\[
\sqrt{s} \approx \sqrt{s_0} \left(1 + \frac{\delta s}{2s_0}\right)
\]

Therefore:

\[
\sinh(\sqrt{s} \cdot x) \approx \sinh(\sqrt{s_0} \cdot x) + \cosh(\sqrt{s_0} \cdot x) \cdot \frac{x \cdot \delta s}{2\sqrt{s_0}}
\]

The denominator also has \( \cosh(\sqrt{s} \cdot l) \):

\[
\cosh(\sqrt{s} \cdot l) \approx \cosh(\sqrt{s_0} \cdot l) + \sinh(\sqrt{s_0} \cdot l) \cdot \frac{l \cdot \delta s}{2\sqrt{s_0}}
\]

### Step 4: Compute the residue

Now, let's find the residue at \( s = s_0 \):

The residue at a simple pole \( s_0 \) is:

\[
\text{Res}(f, s_0) = \lim_{s \to s_0} (s - s_0)f(s)
\]

Substituting \( s = s_0 + \delta s \) into \( f(s) \):

\[
f(s_0 + \delta s) \approx \frac{\cosh(\sqrt{s_0} \cdot x) \cdot \exp(s_0 \cdot t)}{s_0 \cdot \frac{l}{2\sqrt{s_0}} \cdot \cosh(\sqrt{s_0} \cdot l) \cdot \sqrt{s_0}} \cdot \delta s
\]

Simplifying:

\[
f(s) \approx \frac{2 \cdot \cosh(\sqrt{s_0} \cdot x) \cdot \exp(s_0 \cdot t)}{l \cdot s_0 \cdot \cosh(\sqrt{s_0} \cdot l)} \cdot \frac{1}{\delta s}
\]

The residue is:

\[
\text{Res}(f, s_0) = \frac{2 \cdot \cosh\left(i \cdot \frac{(2n - 1)\pi x}{2l}\right) \cdot \exp\left(-\left(\frac{(2n - 1)\pi}{2l}\right)^2 \cdot t\right)}{l \cdot \left(-\left(\frac{(2n - 1)\pi}{2l}\right)^2\right) \cdot \cosh\left(i \cdot \frac{(2n - 1)\pi}{2}\right)}
\]

Since \( \cosh(i\theta) = \cos(\theta) \), we can simplify:

\[
\text{Res}(f, s_0) = \frac{2 \cdot \cos\left(\frac{(2n - 1)\pi x}{2l}\right) \cdot \exp\left(-\left(\frac{(2n - 1)\pi}{2l}\right)^2 \cdot t\right)}{l \cdot \left(-\left(\frac{(2n - 1)\pi}{2l}\right)^2\right) \cdot \cos\left(\frac{(2n - 1)\pi}{2}\right)}
\]

Simplify the constants:

\[
\text{Res}(f, s_0) = \frac{-8l \cdot \cos\left(\frac{(2n - 1)\pi x}{2l}\right) \cdot \exp\left(-\left(\frac{(2n - 1)\pi}{2l}\right)^2 \cdot t\right)}{(2n - 1)^2 \pi^2 \cdot \cos\left(\frac{(2n - 1)\pi}{2}\right)}
\]

### Final Answer:

The residue at \( s = -\left(\frac{(2n - 1)\pi}{2l}\right)^2 \) is:

\[
\text{Res}(f, s_0) = \frac{-8l \cdot \cos\left(\frac{(2n - 1)\pi x}{2l}\right) \cdot \exp\left(-\left(\frac{(2n - 1)\pi}{2l}\right)^2 \cdot t\right)}{(2n - 1)^2 \pi^2 \cdot \cos\left(\frac{(2n - 1)\pi}{2}\right)}
\]

Would you like more details or have any questions?

### Related Questions:

1. What is the significance of residues in evaluating complex integrals?
2. How does the residue theorem apply to real-world problems?
3. What is the relationship between poles and residues in the complex plane?
4. Can you derive the Laurent series for this function around its pole?
5. How does the cosine function affect the residue calculation in this context?

### Tip:

Always check the periodicity of trigonometric and hyperbolic functions when dealing with residues, as they can simplify or complicate the residue calculation depending on the point of interest.

Explore how to compute the residue of a complex function at the point s = -(((2*n - 1)*Pi)/(2*l))^2. This detailed guide covers the application of the Residue Theorem, evaluating singularities, and simplifying complex expressions involving hyperbolic and trigonometric functions. Suitable for college-level students and beyond.

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