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xleb123 · Accepted Answer

To obtain the inverse Laplace transform for

\frac{1}{(s - 1) (s^{2} + 1)}

using the convolution theorem, we can express the given function as a product of two Laplace transforms.
Let's rewrite the function as

\frac{1}{s - 1} \cdot \frac{1}{s^{2} + 1}

.
Using the Laplace transform table, we know that the inverse Laplace transform of

\frac{1}{s - 1}

is

e^{t}

, and the inverse Laplace transform of

\frac{1}{s^{2} + 1}

is

\sin (t)

.
According to the convolution theorem, the inverse Laplace transform of the product of two functions can be found by convolving their individual inverse Laplace transforms. Therefore, the inverse Laplace transform of

\frac{1}{(s - 1) (s^{2} + 1)}

is given by the convolution of

e^{t}

and

\sin (t)

.
The convolution integral is defined as:

(f * g) (t) = \int_{0}^{t} f (t - τ) g (τ) d τ

In our case, we have

f (t) = e^{t}

and

g (t) = \sin (t)

.
Therefore, the inverse Laplace transform of

\frac{1}{(s - 1) (s^{2} + 1)}

is given by:

(e^{t} * \sin (t)) (t) = \int_{0}^{t} e^{t - τ} \sin (τ) d τ

Simplifying the integral and evaluating it, we can find the inverse Laplace transform.

\int_{0}^{t} e^{t - τ} \sin (τ) d τ = - e^{t - τ} \cos (τ) |_{0}^{t} + \int_{0}^{t} e^{t - τ} \cos (τ) d τ

Simplifying further, we get:

- (e^{t} \cos (t) - \cos (0)) + \int_{0}^{t} e^{t - τ} \cos (τ) d τ

= - e^{t} \cos (t) + 1 + \int_{0}^{t} e^{t - τ} \cos (τ) d τ

The remaining integral can be solved using integration techniques.
Therefore, the inverse Laplace transform of

\frac{1}{(s - 1) (s^{2} + 1)}

is given by:

- e^{t} \cos (t) + 1 + \int_{0}^{t} e^{t - τ} \cos (τ) d τ

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