Problem 19
Question
(a) From formula (7) in the text with \(x=2,\) we have $$\frac{1}{2}=\frac{2}{\pi} \int_{0}^{\infty} \frac{\sin \alpha \cos \alpha}{\alpha} d \alpha=\frac{1}{\pi} \int_{0}^{\infty} \frac{\sin 2 \alpha}{\alpha} d \alpha$$ If we let \(\alpha=x\) we obtain $$\int_{0}^{\infty} \frac{\sin 2 x}{x} d x=\frac{\pi}{2}$$ (b) If we now let \(2 x=k t\) where \(k > 0\), then \(d x=(k / 2) d t\) and the integral in part (a) becomes
Step-by-Step Solution
Verified Answer
The integral becomes \(\int_{0}^{\infty} \frac{\sin(kt)}{t} dt = \frac{\pi}{2}\).
1Step 1: Substitute α with x in the Original Formula
Start with the given integral \[ \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin 2 \alpha}{\alpha} d \alpha = \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin 2 x}{x} d x \]and recognize that the result of the integral is \(\frac{\pi}{2}\). This step confirms that \(\int_{0}^{\infty} \frac{\sin 2 x}{x} dx = \frac{\pi}{2}\).
2Step 2: Make a New Substitution
For the transformation, substitute \(2x = kt\). This means \(x = \frac{kt}{2}\) and therefore, the differential \(dx\) becomes \(\frac{k}{2} dt\). Rewriting the expression for \(\sin 2x\) gives \(\sin(kt)\).
3Step 3: Adjust Limits and Differential
Substitute into the integral: \(\int_{0}^{\infty} \frac{\sin 2 x}{x} dx\to \int_{0}^{\infty} \frac{\sin(kt)}{\frac{kt}{2}} \cdot \frac{k}{2} dt\). The \(\frac{k}{2}\) from the substitution multiplies the integral, simplifying the expression.
4Step 4: Simplify and Solve the New Integral
Simplify the integral:\[\int_{0}^{\infty} \frac{\sin(kt) k}{k t} dt = \int_{0}^{\infty} \frac{\sin(kt)}{t} dt\]Since the substitution preserves the form of a standard integral, the value of \(\int_{0}^{\infty} \frac{\sin(kt)}{t} dt\) simplifies to \(\frac{\pi}{2}\) because it's similar to the original integral when considering the substitution.
Key Concepts
Integral TransformationSubstitution MethodTrigonometric IntegralsLimit Evaluation
Integral Transformation
Integral transformation helps in solving complex integrals by converting them into a simpler form. In calculus, this concept is particularly useful in transforming terms to align with known integral forms or rules. It allows you to replace one variable with another, leading to a more convenient representation.
The main goal of integral transformation is to make the integral look like something you've already established or solved previously. For example, let's look at the transformation we did by changing variables to make the form more recognizable. This is done by using strategies like substitution or parts of the integral to link it to a known integral solution.
When dealing with improper integrals, as seen in our exercise, we made the substitution that helps match the known integral of \ \(\sin 2x/x\ \). This manipulation helps us efficiently solve the problem with a known result \(\frac{\pi}{2}\), showcasing the power of integral transformation in mathematical simplification.
The main goal of integral transformation is to make the integral look like something you've already established or solved previously. For example, let's look at the transformation we did by changing variables to make the form more recognizable. This is done by using strategies like substitution or parts of the integral to link it to a known integral solution.
When dealing with improper integrals, as seen in our exercise, we made the substitution that helps match the known integral of \ \(\sin 2x/x\ \). This manipulation helps us efficiently solve the problem with a known result \(\frac{\pi}{2}\), showcasing the power of integral transformation in mathematical simplification.
Substitution Method
The substitution method is a powerful technique to simplify complex integrals. This approach involves substituting a part of the integral with a new variable to help simplify calculations.
Take our example with the substitution \(2x = kt\). We rewrite the variable \(x\) in terms of \(t\) and adjust the differential \(dx\) to \(\frac{k}{2} dt\). This substitution changes the form of the integral, making it easier to solve by linking it to a known form.
Substitution has certain steps: determining the part of the integral that can be substituted, rewriting the differential according to the new variable, and then changing the limits of integration if definite. It makes solving easier by reducing the integral to a familiar form or standard integral.
Take our example with the substitution \(2x = kt\). We rewrite the variable \(x\) in terms of \(t\) and adjust the differential \(dx\) to \(\frac{k}{2} dt\). This substitution changes the form of the integral, making it easier to solve by linking it to a known form.
Substitution has certain steps: determining the part of the integral that can be substituted, rewriting the differential according to the new variable, and then changing the limits of integration if definite. It makes solving easier by reducing the integral to a familiar form or standard integral.
Trigonometric Integrals
Trigonometric integrals are integrals involving trigonometric functions such as sine and cosine. These often appear in problems involving waves, oscillations, and more.
In our context, we dealt with the integral of \(\sin(2x)\). The aim was to manage the trigonometric function within the integral to solve effectively. Certain formulas and identities can simplify these integrals, such as the famous formula for \(\int \sin(nx)/x\ dx\) which often appears in Fourier analysis.
Trigonometric integrals require manipulating the trigonometric functions, sometimes using identities or substitutions, to bring them to a form where they can be easily integrated or compared with known results like we've seen. Recognizing these standard integral forms can greatly aid in quick evaluations.
In our context, we dealt with the integral of \(\sin(2x)\). The aim was to manage the trigonometric function within the integral to solve effectively. Certain formulas and identities can simplify these integrals, such as the famous formula for \(\int \sin(nx)/x\ dx\) which often appears in Fourier analysis.
Trigonometric integrals require manipulating the trigonometric functions, sometimes using identities or substitutions, to bring them to a form where they can be easily integrated or compared with known results like we've seen. Recognizing these standard integral forms can greatly aid in quick evaluations.
Limit Evaluation
Limit evaluation is integral (pun intended!) in dealing with improper integrals. These occur when the limits of integration are infinite or the integrand becomes infinite within the limits.
In the solution, the limits are from 0 to \(\infty\), requiring precise handling. We solve these by usually finding a sort of limit of the integral's behavior as it approaches infinity or another point that causes undefined behavior.
Proper handling means evaluating the integral close to these problematic points, like zero or infinity, ensuring convergence or resolving divergences. It also involves knowing how integrals behave under limits, allowing you to find values definitively, like we've calculated \(\int_0^{\infty} \sin(kt)/t\ dt = \pi/2\). This understanding and calculation ensure that despite the improper design, the integration provides reliable results.
In the solution, the limits are from 0 to \(\infty\), requiring precise handling. We solve these by usually finding a sort of limit of the integral's behavior as it approaches infinity or another point that causes undefined behavior.
Proper handling means evaluating the integral close to these problematic points, like zero or infinity, ensuring convergence or resolving divergences. It also involves knowing how integrals behave under limits, allowing you to find values definitively, like we've calculated \(\int_0^{\infty} \sin(kt)/t\ dt = \pi/2\). This understanding and calculation ensure that despite the improper design, the integration provides reliable results.
Other exercises in this chapter
Problem 15
We use the Fourier sine transform with respect to \(x\) to obtain $$U(\alpha, y)=c_{1} \cosh \alpha y+c_{2} \sinh \alpha y.$$ The transforms of \(u(x, 0)=f(x)\)
View solution Problem 18
We solve the three boundary-value problems: Using separation of variables we find the solution of the first problem is $$u_{1}(x, y)=\sum_{n=1}^{\infty} A_{n} e
View solution Problem 20
Transforming the partial differential equation gives $$k \frac{d^{2} U}{d x^{2}}-s U=-\frac{r}{s}$$ Using undetermined coefficients we obtain $$U(x, s)=c_{1} e^
View solution Problem 21
The solution of $$\frac{d^{2} U}{d x^{2}}-s U=-u_{0}-u_{0} \sin \frac{\pi}{L} x$$ is $$U(x, s)=c_{1} \cosh (\sqrt{s} x)+c_{2} \sinh (\sqrt{s} x)+\frac{u_{0}}{s}
View solution