Question

\(E_{n}(x)=\int_{1}^{\infty} \frac{e^{-x t}}{t^{n}} d t, n=0,1,2, \cdots,\) and \(\operatorname{Ei}(x)=\int_{-\infty}^{x} \frac{e^{t}}{t} d t,\) and other similar integrals are called exponential integrals. By making appropriate changes of variable, show that (a) \(\quad E_{1}(x)=\int_{x}^{\infty} \frac{e^{-t}}{t} d t\) (b) \(\operatorname{Ei}(x)=-\int_{-x}^{\infty} \frac{e^{-t}}{t} d t\) (c) \(\quad E_{1}(x)=-\mathrm{Ei}(-x)\) (d) \(\int_{0}^{x} \frac{e^{1 / t}}{t} d t=E_{1}(-1 / x)\)

Short answer

1) \[ E_1(x)=\int_{x}^{\infty} \frac{e^{-t}}{t} dt \] 2) \[ \operatorname{Ei}(x)=-\int_{-x}^{\infty} \frac{e^{-t}}{t} dt \] 3) \[ E_1(x)=-\operatorname{Ei}(-x) \] 4) \[ \int_{0}^{x} \frac{e^{1/t}}{t} dt=E_1(-1/x) \]}

Step-by-step solution

Derive part (a): Change of variable for \(E_{1}(x)\)

Start with the given integral definition for \(E_n(x)\) where \(n = 1\). Use the given integration formula \[ E_1(x)=\int_{1}^{\infty} \frac{e^{-xt}}{t} dt \]Introduce a change of variable: let \(u = xt\), therefore \(du = x dt\) and \(dt = \frac{du}{x}\). When \(t \to 1\), \(u \to x\) and when \(t \to \infty\), \(u \to \infty\). Substitute these into the integral,\[ E_1(x)=\int_{x}^{\infty} \frac{e^{-u}}{u} du \]Thus, \[ E_1(x)=\int_{x}^{\infty} \frac{e^{-t}}{t} dt \]

Derive part (b): Change of variable for \(Ei(x)\)

Start with the integral definition of the exponential integral \[ \operatorname{Ei}(x)=\int_{-\infty}^{x} \frac{e^{t}}{t} dt \]Introduce a change of variable: let \(u = -t\) so that \(du = -dt\), therefore \(dt = -du\). When \(t \to -\infty\), \(u \to \infty\) and when \(t \to x\), \(u \to -x\). Substitute these into the integral,\[ \operatorname{Ei}(x)=\int_{\infty}^{-x} \frac{e^{-u}}{-u} du \]To make the limits of integration go from \(-x\) to \(\infty\), rewrite the integral as,\[ \operatorname{Ei}(x)=-\int_{-x}^{\infty} \frac{e^{-u}}{u} du \]Thus, \[ \operatorname{Ei}(x)=-\int_{-x}^{\infty} \frac{e^{-t}}{t} dt \]

Derive part (c): Relationship between \(Ei(x)\) and \(E_1(x)\)

Using the results from part (a) and (b):From part (b): \[ \operatorname{Ei}(x)=-\int_{-x}^{\infty} \frac{e^{-t}}{t} dt \]Substitute \(x = -x\),\[ \operatorname{Ei}(-x)=-\int_{x}^{\infty} \frac{e^{-t}}{t} dt \]According to part (a):\[ \int_{x}^{\infty} \frac{e^{-t}}{t} dt=E_1(x) \]Combining these results, we get,\[ E_1(x)=-\operatorname{Ei}(-x) \]

Derive part (d): Change of variable for \(\int_{0}^{x} \frac{e^{1/t}}{t} dt\)

Start with the given integral and make the variable substitution \(u = 1/t\). Then \(du = -1/t^2 dt\), and thus \(dt = -1/u^2 du\).The integral becomes:\[ \int_{0}^{x} \frac{e^{1/t}}{t} dt \rightarrow \int_{\infty}^{1/x} \frac{e^{u}}{1/u} \left( -{1}/{u^2}\, du \right) \]Simplify the integrand:\[ -\int_{1/x}^{\infty} \frac{e^{u}}{u} du \]Recognize that this integral matches the definition of \(E_{1}(y)\) where \(y = -1/x\):\[ -\int_{1/x}^{\infty} \frac{e^{u}}{u} du = E_1(-1/x) \]

Key concepts

change of variables in integrals

Changing variables in integrals is a technique used to simplify the process of integration. By substituting a variable with another one, you can transform the integral into a more familiar or solvable form. For instance, when integrating an exponential function, this method often simplifies the integral and makes it easier to handle. To make a change of variables:
- Identify a substitution that simplifies the integrand.
- Compute the differential change: if you substitute with a new variable, calculate the derivative to find the new differential element (e.g., if you let \( u = xt \), then \( du = x dt \)).
- Adjust the limits of integration: translate the bounds of the original variable to the new variable.
- Substitute and simplify. This change often transforms a complex problem into a standard form that you can solve using basic techniques.

definite integrals

Definite integrals have specified limits of integration and provide the net area between the graph of a function and the x-axis, within these limits. They’re essential for finding quantities such as area, volume, and total change. When evaluating a definite integral:
- Find the antiderivative of the integrand.
- Apply the Fundamental Theorem of Calculus by evaluating the antiderivative at the upper limit and subtracting its value at the lower limit.
- Consider the geometric interpretation of the area under the curve.
In our example, the integral \(E_{1}(x)\) from the exercise involves transforming a definite integral through substitution to find simpler forms that still respect the defined limits.

exponential functions

Exponential functions are fundamental in many areas of mathematics and science, characterized by a constant base raised to a variable exponent. They are written in the form \(e^{x}\) for the natural exponential function. These functions have distinctive properties:
- Rapid growth or decay: They grow or decay exponentially, faster than any polynomial.
- The derivative of \(e^{x}\) is \(e^{x}\), and the integral of \(e^{x}\) is also \(e^{x}\).

In the context of integrals, exponential functions often simplify when integrating due to these properties. For instance, integrals like \( \frac{e^{-xt}}{t^{n}} \) can be transformed into simpler forms using substitutions, leveraging the nature of exponential growth and decay.

integral transformations

Integral transformations involve techniques to convert one integral into another form that is easier to solve. Techniques like substitution and integration by parts fall under this category. The key steps to perform an integral transformation include:
- Selecting a suitable substitution that makes the integral manageable.
- Rewriting the integral in terms of the new variable.
- Simplifying and solving the integral.
In our exercise, we applied transformations to solve exponential integrals. By choosing substitutions wisely, we simplified complex integrals into standard forms, demonstrating the power of integral transformations to tackle otherwise difficult calculus problems.