Problem 87 - Calculus Single Variable

Question

The Fresnel sine and Fresnel cosine functions, denoted by FresnelS and FresnelC, are important in the theory of optics. They are defined by $$ \text { FresnelS }(x)=\int_{0}^{x} \sin \left(\frac{\pi}{2} t^{2}\right) d t $$ and $$ \text { FresnelC }(x)=\int_{0}^{x} \cos \left(\frac{\pi}{2} t^{2}\right) d t $$ a. Integrate by parts to calculate \(\int\) FresnelS \((x) d x\). (Your answer will involve FresnelS \((x)\) and elementary functions.) b. Integrate by parts to calculate \(\int\) Fresnel \(\mathrm{C}(x) d x\). (Your answer will involve FresnelC(x) and elementary functions.) c. Integrate by parts to calculate \(\int 2 x\) FresnelS \((x) d x\). (Your answer will involve both Fresnel functions and elementary functions.) d. Integrate by parts to calculate \(\int 2 x\) Fresnel \(\mathrm{C}(x) d x\). (Your answer will involve both Fresnel functions and elementary functions.)

Step-by-Step Solution

Answer

Integration by parts applied to each yields different integrals involving FresnelS, FresnelC, sine and cosine terms.

Step 1: Understand the Problem

We need to use integration by parts to calculate four different integrals involving the Fresnel sine and cosine functions. These integrals involve their derivatives which were given in the problem statement as functions of sine and cosine integrals.

Step 2: Part a - Determine u and dv

For \( \int \text{FresnelS}(x) \, dx \), apply integration by parts where \( u = \text{FresnelS}(x) \) and \( dv = dx \). Thus, \( du = \sin\left(\frac{\pi}{2} x^2\right) \, dx \) and \( v = x \).

Step 3: Part a - Apply Integration by Parts Formula

Using \( \int u \, dv = uv - \int v \, du \), get:\[\int \text{FresnelS}(x) \, dx = x \cdot \text{FresnelS}(x) - \int x \sin\left(\frac{\pi}{2} x^2\right) \, dx\]

Step 4: Part b - Determine u and dv

For \( \int \text{FresnelC}(x) \, dx \), choose \( u = \text{FresnelC}(x) \) and \( dv = dx \). Here, \( du = \cos\left(\frac{\pi}{2} x^2\right) \, dx \) and \( v = x \).

Step 5: Part b - Apply Integration by Parts Formula

Apply the formula for integration by parts:\[\int \text{FresnelC}(x) \, dx = x \cdot \text{FresnelC}(x) - \int x \cos\left(\frac{\pi}{2} x^2\right) \, dx\]

Step 6: Part c - Determine u and dv

For \( \int 2x \cdot \text{FresnelS}(x) \, dx \), set \( u = \text{FresnelS}(x) \) and \( dv = 2x \, dx \). Thus, \( du = \sin\left(\frac{\pi}{2} x^2\right) \, dx \) and \( v = x^2 \).

Step 7: Part c - Apply Integration by Parts Formula

The integration by parts yields:\[\int 2x \cdot \text{FresnelS}(x) \, dx = x^2 \cdot \text{FresnelS}(x) - \int x^2 \sin\left(\frac{\pi}{2} x^2\right) \, dx\]

Step 8: Part d - Determine u and dv

For \( \int 2x \cdot \text{FresnelC}(x) \, dx \), let \( u = \text{FresnelC}(x) \) and \( dv = 2x \, dx \). Then, \( du = \cos\left(\frac{\pi}{2} x^2\right) \, dx \) and \( v = x^2 \).

Step 9: Part d - Apply Integration by Parts Formula

Now, apply integration by parts:\[\int 2x \cdot \text{FresnelC}(x) \, dx = x^2 \cdot \text{FresnelC}(x) - \int x^2 \cos\left(\frac{\pi}{2} x^2\right) \, dx\]

Step 10: Conclusion

Each integral is calculated by integrating by parts, simplifying to integrals involving the Fresnel functions with sine and cosine terms inside the integrals.

Key Concepts

Fresnel Sine Function

The Fresnel Sine Function, represented as \( \text{FresnelS}(x) \), is a special function that is crucial in the field of optics. It is defined by the integral:

\[\text{FresnelS}(x) = \int_{0}^{x} \sin\left(\frac{\pi}{2} t^2\right) dt\]
This function arises in problems involving wave propagation and diffraction. The integral calculates the accumulation of a sine wave with a variable frequency, which is affected by the square of the variable \( t \).

• The Fresnel Sine Function does not have a simple closed-form expression, making it an interesting and complex aspect of mathematical analysis.
• In optics, it aids in calculating light patterns, especially how light waves bend and spread when passing through lenses or small apertures.
• Integration by parts is often used with these functions to find their antiderivatives, as illustrated in the exercise above.

Fresnel Cosine Function

The Fresnel Cosine Function is similarly significant in the field of optics. It is defined as follows:

\[\text{FresnelC}(x) = \int_{0}^{x} \cos\left(\frac{\pi}{2} t^2\right) dt\]
Here, the focus is on the cosine wave. This function, together with the Fresnel Sine Function, models complex wave behaviors such as interference and diffraction patterns.

• It is used in scenarios where phase differences create intricate patterns over time or space.
• The Fresnel Cosine Function, like its sine counterpart, does not resolve into elementary functions.
• In mathematical terms, it is central to calculating integrals involving variable frequency cosine waves, often requiring specific numerical or approximation methods.
• As demonstrated in the exercise, integration by parts helps decompose these functions into more manageable expressions.

Optics

Optics is the branch of physics that deals with the behavior and properties of light. It encompasses both practical applications and theoretical constructs such as the Fresnel functions.

• Optical physics helps in understanding how light interacts with matter and how it propagates through different media.
• The Fresnel Sine and Cosine Functions are important in analysing wave diffraction and interference, which describe how light waves overlap to create various patterns.
• These functions assist in predicting and explaining patterns that occur when light waves pass through lenses or across small openings.
• In devices like glasses, cameras, and microscopes, these principles come into play to enhance images or focus light precisely.

Elementary Functions

Elementary functions include the fundamental functions in mathematics like polynomials, exponential functions, trigonometric functions, and their inverses. They form the building blocks for more complex functions such as Fresnel functions.

• Elementary functions are characterized by their simple formulas which can be expressed using basic mathematical operations.
• In contrast, Fresnel functions cannot be expressed in terms of these elementary functions, which makes them interesting subjects for study.
• In the exercise, we saw that the results of integrating Fresnel functions often yield new expressions that include elementary functions.
• Understanding the difference between these classes of functions is crucial when tackling integration problems in advanced mathematical contexts.