Question

Let \(n\) be a positive integer. Using integration by parts, establish the reduction formula $$ \int t^{n} e^{-s t} d t=-\frac{t^{n} e^{-s t}}{s}+\frac{n}{s} \int t^{n-1} e^{-s t} d t, \quad s>0 . $$

Short answer

Question: Establish the reduction formula for the integral of the function involving a power of the variable multiplied by an exponential term $t^n e^{-st} dt$, where $s > 0$. Answer: $\int t^{n} e^{-s t} d t=-\frac{t^{n} e^{-s t}}{s}+\frac{n}{s} \int t^{n-1} e^{-s t} d t, \quad s>0 $.

Step-by-step solution

Integration by parts formula

Recall the integration by parts formula, which states: $$ \int u \, dv = uv - \int v \, du, $$ where \(u\) is the function to differentiate, and \(v\) is the function to integrate. In our problem, we have \(u=t^n\) and \(dv=e^{-st} dt\). Now we need to find \(du\) and \(v\).

Find du and v

First, we find the derivative of \(u\) with respect to \(t\): $$ du = \frac{d}{dt}(t^n) = n t^{n-1} dt. $$ Next, we find the integral of \(dv\) with respect to \(t\): $$ v=\int e^{-st} dt = -\frac{1}{s} e^{-st} + C. $$ Note that, since we are looking for a general formula., the constant of integration, \(C\), does not contribute to the final result. Therefore, we only consider the non-constant part of \(v\), which is \(-\frac{1}{s} e^{-st}\).

Apply integration by parts formula

Now we can apply the integration by parts formula using the expressions we found for \(u\), \(du\), \(v\), and \(dv\): $$ \int t^n e^{-st} dt = \left( -\frac{t^n}{s} e^{-st} \right) - \int \left( -\frac{n t^{n-1}}{s} e^{-st} \right) dt. $$

Simplify the expression

Finally, we simplify the expression to get the desired reduction formula: $$ \int t^n e^{-st} dt = -\frac{t^n e^{-st}}{s} + \frac{n}{s} \int t^{n-1} e^{-st} dt. $$ As a result, we have established the reduction formula: $$ \int t^{n} e^{-s t} d t=-\frac{t^{n} e^{-s t}}{s}+\frac{n}{s} \int t^{n-1} e^{-s t} d t, \quad s>0 . $$

Key concepts

Integration by Parts

Integration by parts is a strategy specific for solving complex integrals that are difficult to evaluate by standard approaches. It transforms the integral into a different form, which may be more easily calculated.

The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

In this formula, you need to identify parts of the integrand as \( u \) and \( dv \). Then, compute \( du \) (the derivative of \( u \)) and \( v \) (the integral of \( dv \)).

The choice of \( u \) and \( dv \) is crucial. A typical rule is to let \( u \) be a function that becomes simpler when you differentiate it, and \( dv \) to be a function easily integrable. This is a helpful tool in calculus, especially when dealing with polynomial and exponential functions.

Exponential Function

The exponential function refers to functions written in the form \( e^x \), where \( e \) is a mathematical constant approximately equal to 2.71828. In the context of calculus, such functions have unique properties, especially when it comes to differentiation and integration.

The derivative of \( e^{ax} \), where \( a \) is a constant, is \( ae^{ax} \). This is due to the constant rate of growth of exponential functions. Conversely, the integral of \( e^{ax} \) is \( \frac{1}{a}e^{ax} + C \), where \( C \) is the constant of integration.

This consistent behavior makes exponential functions favorable in calculations. In the problem involving the reduction formula, the exponential term \( e^{-st} \) was chosen for \( dv \) because its integration is direct, leading to a manageable expression for \( v \).

Inductive Process

An inductive process is a method of reasoning that involves forming generalizations based on individual instances or patterns. In mathematics, it is often used in proofs by mathematical induction and in establishing formulas.

In the context of the reduction formula, an inductive approach might be used to verify that the formula holds for any positive integer \( n \). Normally, this starts by proving the formula for a base case, such as \( n = 1 \). Next, one assumes it holds for \( n = k \) and demonstrates it still holds for \( n = k+1 \).

This method of logical stepping ensures generality and applicability of mathematical results across a range of values, validating the reduction formula for diminishing powers of \( t \) as needed.

Derivative and Integral Calculation

Calculating derivatives and integrals involves different rules and formulas, which are essential parts of calculus. These calculations are used to determine the rate of change and the cumulative effect of functions, respectively.

In derivatives, understanding the power rule is crucial. For \( t^n \), the derivative is found using \( \frac{d}{dt}(t^n) = nt^{n-1} \). This notation reflects how powers decrease as being differentiated.

For integrals, especially involving exponential functions, remember the rule for integrating \( e^{-ax} \), which states that the integral is \( -\frac{1}{a}e^{-ax} + C \). These calculations allow for transition between differentiating polynomial and integrating exponential functions.

In solving problems like reduction formulas, these concepts help establish connections between different parts of integrals, ultimately simplifying the solution path.