Problem 97 - Calculus and its applications

Question

Evaluate. $$ \int \frac{e^{1 / t}}{t^{2}} d t $$

Step-by-Step Solution

Answer

\( e^{1/t} \left(t + \ln |t|\right) + C \)

Step 1: Identify the Substitution

Notice that the integrand is of the form \( \frac{e^{1/t}}{t^2} \). A suitable substitution is to let \( u = \frac{1}{t} \) which implies that \( t = \frac{1}{u} \). This means that \( dt = -\frac{1}{u^2} du \).

Step 2: Rewrite the Integral

Substitute the expressions from Step 1 into the integral. We have:\[\int \frac{e^{1/t}}{t^2} \, dt = \int e^u \left(-\frac{1}{u^2}\right) \, du\] which simplifies to:\[-\int \frac{e^u}{u^2} \, du\]

Step 3: Determine if Further Substitution is Needed

The resulting integral is now:\[ -\int \frac{e^u}{u^2} \, du \]This integral requires integration by parts or recognition of a standard form which might involve a special function. However, it is typically considered in terms of the exponential integral function \( \text{Ei}(x) \).

Step 4: Express Using an Exponential Integral

Recognize that the integral can be expressed in terms of an exponential integral function: \[ \int \frac{e^u}{u^2} \, du = -e^u \left(\frac{1}{u} + \ln |u| + C'\right) \]Thus the full solution becomes \[-\left(-\frac{e^u}{u} - e^u \ln |u| + C'\right) = \frac{e^u}{u} + e^u \ln |u| + C' \]

Step 5: Substitute Back in Terms of t

Replace \( u \) back with \( \frac{1}{t} \).Thus, \[ \int \frac{e^{1/t}}{t^2} \, dt = \frac{e^{1/t}}{1/t} + e^{1/t} \ln |1/t| + C \]Simplifying gives:\[ t e^{1/t} + e^{1/t} \ln |t| + C \]

Step 6: Write the Final Solution

The evaluated integral is given as:\[\int \frac{e^{1/t}}{t^2} \, dt = e^{1/t} \left(t + \ln |t|\right) + C\]

Key Concepts

Exponential Integral

The Exponential Integral, often denoted as \( \text{Ei}(x) \), is a special function that appears frequently in calculus, especially when dealing with integrals involving exponential functions divided by polynomials or powers of \( x \). Unlike elementary functions, which are defined through basic algebraic operations, the exponential integral requires integration and manipulation of more complex expressions. To give a bit more clarity, the Exponential Integral \( \text{Ei}(x) \) is specifically defined through an integral itself, namely:

• \( \text{Ei}(x) = \int \frac{e^x}{x} \, dx \)

This is important because it cannot be expressed in terms of basic functions like polynomials, radicals, or even simple exponential functions. It emerges in solving integrals similar to the one we're considering, where a straightforward antiderivative might not exist. The Exponential Integral function is essential in tackling these kinds of integrals neatly, allowing mathematicians and students alike to work with problems that involve exponential decay or growth, particularly when these are intertwined with reciprocal functions.

Integration by Substitution

Integration by substitution is a technique that simplifies an integral by changing its variables. The method is essentially the reverse of the chain rule for differentiation and is one of the fundamental strategies in integral calculus.In the exercise, the given integral \( \int \frac{e^{1/t}}{t^2} \, dt \) is simplified using this technique. The substitution begins with setting the inner function or the complex part equal to a new variable. For instance:

• Let \( u = \frac{1}{t} \), which then makes \( t = \frac{1}{u} \).
• Taking the derivative gives \( dt = -\frac{1}{u^2} \, du \).

By substituting these into the original integral, the form becomes more straightforward. Substitution is about spotting how one piece of an integral can substitute a more complex part, therefore simplifying the solution:

• The original integral was transformed to \(-\int \frac{e^u}{u^2} \, du \).

Understanding how and why we choose a particular substitution is key to mastering this method, as it allows us to solve otherwise complicated integrals by simplifying them to a recognizable form.

Integration by Parts

Integration by Parts is another powerful technique in calculus, stemming from the product rule for differentiation. It provides a way to integrate products of functions and is particularly useful for integrals involving polynomials multiplied by trig, exponential, or logarithmic functions.In the context of this solution, once the substitution refined the integral to \(-\int \frac{e^u}{u^2} \, du \), the expression hints at needing further manipulation. The connection is through the recognition that certain integrals can be expressed and solved through the exponential integral function. This may not always be solved explicitly through integration by parts alone. However, the integration by parts formula is generally:

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

Each integral is analyzed for suitability to fit this format, ensuring a path forward toward simplicity.In some cases, as seen in the exercise, integration by parts transitions into recognizing or categorizing it as a special function, thus avoiding a lengthy manual integration process.Whether directly applied or indirectly by transitioning to standard forms, integration by parts is instrumental in expanding your toolbox when solving integrals related to mathematical and physical problems.