Why is the integral from 0 to -infinity and not +infinity. Can you please explain it to me? Thank you
Why is the integral from 0 to -infinity and not +infinity. Can you please explain it to me? Thank you
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Concept explainers
Riemann Sum
Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus.
Riemann Integral
Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.
Question
Why is the
![### Solution for Advanced Mathematics Problem
In this problem, we are required to evaluate the integral \( I \). The solution involves transforming the integral in polar coordinates and simplifying it using exponential substitution.
#### Step-by-Step Solution:
Given the integral:
\[
I = \iint e^{-a(x^2 + y^2)} \, dx \, dy
\]
Transform to polar coordinates where \( x = r\cos\theta \) and \( y = r\sin\theta \):
\[
I = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-ar^2} r \, dr \, d\theta
\]
Next, use substitution to simplify the integrand. Let:
\[
t = -ar^2 \implies dt = -2ar \, dr \implies r \, dr = \frac{dt}{-2a}
\]
Thus, the integral becomes:
\[
I = \int_{0}^{2\pi} \int_{0}^{-\infty} e^{t} \frac{dt}{-2a} \, d\theta
\]
Separate the variables and simplify:
\[
I = 2\pi \int_{0}^{-\infty} e^t \frac{dt}{-2a}
\]
Simplify further:
\[
I = -\frac{1}{2a} \cdot 2\pi \left[ e^t \right]_{0}^{-\infty}
\]
Evaluate the limit:
\[
I = -\frac{\pi}{a} \left( e^{-\infty} - e^0 \right)
\]
Since \( e^{-\infty} \to 0 \) and \( e^0 = 1 \):
\[
I = -\frac{\pi}{a} (0 - 1)
\]
Finally:
\[
I = \frac{\pi}{a}
\]
Thus, the integral evaluates to:
\[
I = \frac{\pi}{a}
\]
By following these steps, the integral that initially appeared complex due to the exponent and limits becomes manageable and yields a simple and elegant result.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2Fea44dd71-233e-4c1b-bfd7-12f815e979a7%2Fmi4fpg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solution for Advanced Mathematics Problem
In this problem, we are required to evaluate the integral \( I \). The solution involves transforming the integral in polar coordinates and simplifying it using exponential substitution.
#### Step-by-Step Solution:
Given the integral:
\[
I = \iint e^{-a(x^2 + y^2)} \, dx \, dy
\]
Transform to polar coordinates where \( x = r\cos\theta \) and \( y = r\sin\theta \):
\[
I = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-ar^2} r \, dr \, d\theta
\]
Next, use substitution to simplify the integrand. Let:
\[
t = -ar^2 \implies dt = -2ar \, dr \implies r \, dr = \frac{dt}{-2a}
\]
Thus, the integral becomes:
\[
I = \int_{0}^{2\pi} \int_{0}^{-\infty} e^{t} \frac{dt}{-2a} \, d\theta
\]
Separate the variables and simplify:
\[
I = 2\pi \int_{0}^{-\infty} e^t \frac{dt}{-2a}
\]
Simplify further:
\[
I = -\frac{1}{2a} \cdot 2\pi \left[ e^t \right]_{0}^{-\infty}
\]
Evaluate the limit:
\[
I = -\frac{\pi}{a} \left( e^{-\infty} - e^0 \right)
\]
Since \( e^{-\infty} \to 0 \) and \( e^0 = 1 \):
\[
I = -\frac{\pi}{a} (0 - 1)
\]
Finally:
\[
I = \frac{\pi}{a}
\]
Thus, the integral evaluates to:
\[
I = \frac{\pi}{a}
\]
By following these steps, the integral that initially appeared complex due to the exponent and limits becomes manageable and yields a simple and elegant result.
![### Advanced Mathematics: Solving Integral using Polar Coordinates
#### Step 1
We start with the given integral:
\[ I = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-a^2(x^2 + y^2)} \, dx \, dy \quad \text{(i)} \]
By using the polar substitution, we have:
\[
x = r\cos\theta, \quad y = r\sin\theta
\]
Where \(0 \leq r < \infty\) and \(0 \leq \theta < 2\pi\).
The given integral can now be transformed in terms of polar coordinates. Note that:
\[ x^2 + y^2 = r^2 \]
We will also need to transform the infinitesimals: \(dx \, dy\).
Calculating the Jacobian determinant for the transformation:
\[
J = \frac{\partial(x,y)}{\partial(r,\theta)} =
\begin{vmatrix}
\cos\theta & -r\sin\theta \\
\sin\theta & r\cos\theta
\end{vmatrix}
\]
This simplifies to:
\[
J = r \cos^2 \theta + r \sin^2 \theta
= r (\cos^2 \theta + \sin^2 \theta)
= r \cdot 1
= r
\]
Hence the area element transforms as:
\[
dx \, dy = |J| \, dr \, d\theta
= r \, dr \, d\theta
\]
So, we derive:
\[
dx \, dy = r \, dr \, d\theta
\]
#### Step 2
The subsequent steps of the integration and solutions would then follow, typically involving integrating over the angular part and the radial part separately, utilizing the separability in polar coordinates. You would integrate:
\[
I = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} e^{-a^2 r^2} r \, dr
\]
This clarifies the substitution and transformation of the integral from Cartesian to polar coordinates for easier computation.
#### Diagrams Explanation
This section does not contain graphs or diagrams. However, if explaining this step involved a graph, it would typically include:
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2Fea44dd71-233e-4c1b-bfd7-12f815e979a7%2Fgs2a6wt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Advanced Mathematics: Solving Integral using Polar Coordinates
#### Step 1
We start with the given integral:
\[ I = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-a^2(x^2 + y^2)} \, dx \, dy \quad \text{(i)} \]
By using the polar substitution, we have:
\[
x = r\cos\theta, \quad y = r\sin\theta
\]
Where \(0 \leq r < \infty\) and \(0 \leq \theta < 2\pi\).
The given integral can now be transformed in terms of polar coordinates. Note that:
\[ x^2 + y^2 = r^2 \]
We will also need to transform the infinitesimals: \(dx \, dy\).
Calculating the Jacobian determinant for the transformation:
\[
J = \frac{\partial(x,y)}{\partial(r,\theta)} =
\begin{vmatrix}
\cos\theta & -r\sin\theta \\
\sin\theta & r\cos\theta
\end{vmatrix}
\]
This simplifies to:
\[
J = r \cos^2 \theta + r \sin^2 \theta
= r (\cos^2 \theta + \sin^2 \theta)
= r \cdot 1
= r
\]
Hence the area element transforms as:
\[
dx \, dy = |J| \, dr \, d\theta
= r \, dr \, d\theta
\]
So, we derive:
\[
dx \, dy = r \, dr \, d\theta
\]
#### Step 2
The subsequent steps of the integration and solutions would then follow, typically involving integrating over the angular part and the radial part separately, utilizing the separability in polar coordinates. You would integrate:
\[
I = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} e^{-a^2 r^2} r \, dr
\]
This clarifies the substitution and transformation of the integral from Cartesian to polar coordinates for easier computation.
#### Diagrams Explanation
This section does not contain graphs or diagrams. However, if explaining this step involved a graph, it would typically include:
1
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
