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))

To find the residue of the given function at the specified point, \( s = \left(\frac{(2n - 1)\pi}{2l}\right)^2 \), we need to analyze the function and determine if it has a pole at this point, then compute the residue at this pole.

### Given function:

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

### Step 1: Identify the location of the poles

The potential poles of the function are determined by the zeros of the denominator. Let's break down the denominator:

\[
\text{Denominator} = s \cosh(\sqrt{s} \cdot l) \cdot \sqrt{s}
\]

This function can have poles if \( \cosh(\sqrt{s} \cdot l) = 0 \).

### Step 2: Solve for the zeros of \( \cosh(\sqrt{s} \cdot l) \)

We know that the hyperbolic cosine function \( \cosh(z) \) has no real zeros. However, it has complex zeros. Specifically, \( \cosh(z) = 0 \) when:

\[
z = \frac{(2n - 1)\pi i}{2}
\]

So, \( \sqrt{s} \cdot l = \frac{(2n - 1)\pi i}{2} \).

This implies:

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

Square both sides:

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

Thus, the poles of the function are at \( s = -\left(\frac{(2n - 1)\pi}{2l}\right)^2 \).

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

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

Let's rewrite the function by substituting \( s = \left(\frac{(2n - 1)\pi}{2l}\right)^2 \):

\[
f(s) = \frac{\sinh\left(\frac{(2n - 1)\pi x}{2l}\right) \cdot \exp\left(\left(\frac{(2n - 1)\pi}{2l}\right)^2 \cdot t\right)}{\left(\frac{(2n - 1)\pi}{2l}\right)^2 \cosh\left(\frac{(2n - 1)\pi l}{2l}\right) \cdot \frac{(2n - 1)\pi}{2l}}
\]

### Step 4: Compute the residue

The residue of a function \( f(s) \) at a simple pole \( s_0 \) is given by:

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

We need to compute this at \( s = \left(\frac{(2n - 1)\pi}{2l}\right)^2 \):

\[
\text{Res}\left(f, s = \left(\frac{(2n - 1)\pi}{2l}\right)^2 \right) = \lim_{s \to \left(\frac{(2n - 1)\pi}{2l}\right)^2 } \left(s - \left(\frac{(2n - 1)\pi}{2l}\right)^2 \right) \cdot f(s)
\]

However, as identified earlier, the pole is actually at \( s = -\left(\frac{(2n - 1)\pi}{2l}\right)^2 \).

### Conclusion

The function \( f(s) \) has a pole at \( s = -\left(\frac{(2n - 1)\pi}{2l}\right)^2 \), not at \( s = \left(\frac{(2n - 1)\pi}{2l}\right)^2 \). Therefore, the residue at \( s = \left(\frac{(2n - 1)\pi}{2l}\right)^2 \) is **0** because the function has no pole at that point.

Would you like any additional details on this process?

### Related Questions:

1. How do you find the residues of a function at its poles in the complex plane?
2. What are the properties of the hyperbolic cosine function, and how do they relate to complex analysis?
3. Can you explain the significance of poles and residues in contour integration?
4. How do you determine if a function has a simple pole or a higher-order pole?
5. What are some applications of residue calculus in solving real-world problems?

### Tip:

When dealing with residues, always confirm the location of the poles by carefully analyzing the function's denominator and checking for both real and complex solutions.

Learn how to find the residue of a complex function at a specified point using the Residue Theorem and complex analysis techniques. This detailed guide includes step-by-step instructions and examples for determining residues at complex poles in mathematical functions.

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