Problem 89 - Calculus Single Variable

Question

The sine integral Si \((x)\) is defined by $$ \operatorname{Si}(x)=\int_{0}^{x} \frac{\sin (t)}{t} d t $$ (The integrand has a continuous extension at \(0 .)\) a. Integrate by parts to calculate \(\int \operatorname{Si}(x) d x\). (Your answer will involve \(\operatorname{Si}(x)\) and elementary functions.) b. Integrate by parts to calculate \(\int x \operatorname{Si}(x) d x\). (Your answer will involve \(\operatorname{Si}(x)\) and elementary functions.)

Step-by-Step Solution

Answer

a. \(x \operatorname{Si}(x) + \cos(x) + C\); b. \(\frac{x^2}{2} \operatorname{Si}(x) + \frac{1}{2}(x \cos(x) - \sin(x)) + C\)."

Step 1: Integration by Parts for Part a

To integrate \( \int \operatorname{Si}(x) \, dx \), we use integration by parts, which states that \( \int u \, dv = uv - \int v \, du \). Let \( u = \operatorname{Si}(x) \) and \( dv = dx \). Then, \( du = \frac{\sin(x)}{x} \, dx \) (since \( \operatorname{Si}'(x) = \frac{\sin(x)}{x} \)) and \( v = x \). Applying the integration by parts formula gives:\[\int \operatorname{Si}(x) \, dx = x \cdot \operatorname{Si}(x) - \int x \cdot \frac{\sin(x)}{x} \, dx = x \operatorname{Si}(x) - \int \sin(x) \, dx.\]

Step 2: Solving the Remaining Integral for Part a

We continue from where we applied the integration by parts formula. The remaining integral is \( \int \sin(x) \, dx \), which is a standard integral. Recall that:\[\int \sin(x) \, dx = -\cos(x) + C,\] where \( C \) is the constant of integration. Therefore:\[\int \operatorname{Si}(x) \, dx = x \operatorname{Si}(x) + \cos(x) + C.\]

Step 3: Integration by Parts for Part b

For \( \int x \operatorname{Si}(x) \, dx \), again use integration by parts. Let \( u = \operatorname{Si}(x) \) and \( dv = x \, dx \). Then \( du = \frac{\sin(x)}{x} \, dx \) and \( v = \frac{x^2}{2} \). Thus, applying the integration by parts formula gives:\[\int x \operatorname{Si}(x) \, dx = \frac{x^2}{2} \operatorname{Si}(x) - \int \frac{x^2}{2} \cdot \frac{\sin(x)}{x} \, dx = \frac{x^2}{2} \operatorname{Si}(x) - \frac{1}{2} \int x \sin(x) \, dx.\]

Step 4: Solving the Remaining Integral for Part b

Now solve \( \int x \sin(x) \, dx \) using integration by parts again. Let \( u = x \) and \( dv = \sin(x) \, dx \). Then \( du = dx \) and \( v = -\cos(x) \). Consequently:\[\int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx = -x \cos(x) + \sin(x) + C.\]Substitute back into the equation from Step 3:\[\int x \operatorname{Si}(x) \, dx = \frac{x^2}{2} \operatorname{Si}(x) + \frac{1}{2}(x \cos(x) - \sin(x)) + C.\]

Key Concepts

Integration by Parts

Integration by parts is a key method for solving integrals that come in a specific form. It's like the product rule for differentiation but applied in reverse. The formula is given by \[ \int u \, dv = uv - \int v \, du, \]where you choose which part of the integrand to label as \( u \) and which as \( dv \). This means our integral is transformed into a different one that might be simpler to solve. This technique is very handy when dealing with products of functions or functions composed with certain special functions, like the sine integral (Si) function seen here.

• Choice of \( u \) and \( dv \): Selecting \( u \) and \( dv \) is crucial in simplifying the integral. Often, \( u \) is chosen to be the part that becomes simpler when differentiated, and \( dv \) is the part that is easily integrated.
• Using Integration by Parts: For example, when integrating \( \int \operatorname{Si}(x) \, dx \), setting \( u = \operatorname{Si}(x) \) and \( dv = dx \) leverages the known derivative of \( \operatorname{Si}(x) \), which is \( \frac{\sin(x)}{x} \).

Understanding how to apply integration by parts allows you to solve these challenging integrals by breaking them down into simpler components.

Definite Integral

The definite integral is a concept that not only provides the area under a curve but also is fundamental in defining and working with functions such as the sine integral. In relation to this exercise, defining \[ \operatorname{Si}(x) = \int_{0}^{x} \frac{\sin(t)}{t} \, dt, \]means we are looking at an integral from 0 to \( x \) which implies we accumulate the area under the curve of \( \frac{\sin(t)}{t} \) starting from 0 up to some value \( x \).

• Bounds of Integration: The limits of integration (0 and \( x \) in this case) determine the range over which we are considering our function. For definite integrals, these bounds are critical since they give us a single numerical value based on the function's behavior over this interval.
• Continuity Consideration: With the sine integral, it is important that \( \frac{\sin(t)}{t} \) can be continuously extended at \( t=0 \), typically using the fact that \( \lim_{{t \to 0}} \frac{\sin(t)}{t} = 1 \). This ensures that the integral is well-defined.

This approach helps visualize how areas under such curves are linked to functions themselves and why defining such functions (like \( \operatorname{Si}(x) \)) in this manner is both useful and meaningful.

Elementary Functions

Elementary functions are the basic building blocks in calculus, encompassing polynomials, exponential functions, logarithmic functions, and trigonometric functions. In the problem involving the sine integral, we see how these elementary functions play a crucial role when using integration by parts.

• Basic Forms: Trigonometric functions like sine and cosine are quintessential elementary functions, making the integral of \( \sin(x) \) a standard antiderivative. In the solution, integrating \( \sin(x) \) results in \(-\cos(x)\), an example of utilizing elementary functions effectively.


• Combining with Special Functions: When integrating special functions such as \( \operatorname{Si}(x) \), elementary functions often appear in the result. For example, when solving \( \int \operatorname{Si}(x) \, dx \), the resulting solution combines \( \operatorname{Si}(x) \) with \( \cos(x) \), highlighting the interaction between elementary and special functions.


• Understanding Through Simplicity: Recognizing these interactions helps simplify the understanding of complex integrals by reducing them to manipulations of familiar, simple functions.

Grasping the behavior and integration of these elementary functions underpins much of integral calculus, serving as the baseline for approaching and solving more complex problems that involve both simple and special functions.