W.8.8.19 Integration of 2 to \infty

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In summary: Yes. An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration.
  • #1
karush
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$\tiny{w.8.8.19}$
$\textsf{Integration of $2$ to $\infty$}$
$$\displaystyle
I_{19} = \int_{2}^{\infty} \frac{\cos\left({\pi/x}\right)}{{x}^{2}}\,dx $$

$\textit{Not sure how to break this down but it appears the denominator will advance faster than the numerator}$
 
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  • #2
karush said:
$\tiny{w.8.8.19}$
$\textsf{Integration of $2$ to $\infty$}$
$$\displaystyle
I_{19} = \int_{2}^{\infty} \frac{\cos\left({\pi/x}\right)}{{x}^{2}}\,dx $$

$\textit{Not sure how to break this down but it appears the denominator will advance faster than the numerator}$

I would begin by writing:

\(\displaystyle I=\lim_{t\to\infty}\left(\int_2^t \frac{\cos\left(\frac{\pi}{x}\right)}{x^2}\,dx\right)\)

Next, I would use the substitution:

\(\displaystyle u=\frac{1}{x}\implies du=-\frac{1}{x^2}\,dx\)

What does our expression for $I$ look like now?
 
  • #3
MarkFL said:
I would begin by writing:

\(\displaystyle I=\lim_{t\to\infty}\left(\int_2^t \frac{\cos\left(\frac{\pi}{x}\right)}{x^2}\,dx\right)\)

Next, I would use the substitution:

\(\displaystyle u=\frac{1}{x}\implies du=-\frac{1}{x^2}\,dx\)

What does our expression for $I$ look like now?

$\displaystyle I=\lim_{t\to\infty} \left(\int_2^t \cos(\pi u) \,du \right)$

I think this is it?
ok why the $t$ and $u$
 
  • #4
karush said:
$\displaystyle I=\lim_{t\to\infty} \left(\int_2^t \cos(\pi u) \,du \right)$

I think this is it?
ok why the $t$ and $u$

We need to change the limits as well in accordance with the substitution, so we would have:

\(\displaystyle I=\lim_{t\to0}\left(\int_t^{\frac{1}{2}} \cos(\pi u)\,du\right)\)

What do we get when we integrate within the limit?
 
  • #5
MarkFL said:
We need to change the limits as well in accordance with the substitution, so we would have:

\(\displaystyle I=\lim_{t\to0}\left(\int_t^{\frac{1}{2}} \cos(\pi u)\,du\right)\)

What do we get when we integrate within the limit?

well this part would be ...
$\displaystyle\int_0^\frac{1}{2} \cos\left({\pi}u\right)\, du
= \dfrac{\sin\left({\pi}u\right)}{{\pi}}
= \dfrac{1}{{\pi}} $
but would that be the limit also?
 
  • #6
karush said:
well this part would be ...
$\displaystyle\int_0^\frac{1}{2} \cos\left({\pi}u\right)\, du
= \dfrac{\sin\left({\pi}u\right)}{{\pi}}
= \dfrac{1}{{\pi}} $
but would that be the limit also?

Yes, I think the way I would handle it is:

\(\displaystyle I=\lim_{t\to0}\left(\int_t^{\frac{1}{2}} \cos(\pi u)\,du\right)=\lim_{t\to0}\left(\frac{1}{\pi}\int_t^{\frac{1}{2}} \cos(\pi u)\,\pi\cdot du\right)=\lim_{t\to0}\left(\frac{1}{\pi}\left[\sin(\pi u)\right]_t^{\frac{1}{2}}\right)=\lim_{t\to0}\left(\frac{1}{\pi}\left(\sin\left(\frac{\pi}{2}\right)-\sin(t)\right)\right)=\lim_{t\to0}\left(\frac{1}{\pi}\left(1-\sin(t)\right)\right)=\frac{1}{\pi}\)
 
  • #7
is this known as a Improper Integral?
 
  • #8
karush said:
is this known as a Improper Integral?

Yes. An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. :D
 

FAQ: W.8.8.19 Integration of 2 to \infty

1. What does "W.8.8.19 Integration of 2 to \infty" mean?

W.8.8.19 Integration of 2 to \infty refers to a specific mathematical concept known as an improper integral. It involves finding the area under the curve of a function that extends to infinity on the upper limit, in this case from 2 to infinity.

2. How is "W.8.8.19 Integration of 2 to \infty" different from regular integration?

Unlike regular integration, W.8.8.19 Integration of 2 to \infty deals with functions that extend to infinity on the upper limit. This means that the area under the curve cannot be calculated using traditional integration methods and requires a different approach.

3. What are some real-world applications of "W.8.8.19 Integration of 2 to \infty"?

Improper integrals, such as W.8.8.19 Integration of 2 to \infty, have many real-world applications in fields such as physics, engineering, and economics. For example, it can be used to calculate the total distance traveled by an object with a changing velocity or the total profit of a business with fluctuating revenue.

4. How do you solve "W.8.8.19 Integration of 2 to \infty"?

Solving W.8.8.19 Integration of 2 to \infty involves using specific techniques, such as limits and substitution, to manipulate the integral and find a solution. The process can be complex and may require advanced mathematical knowledge and skills.

5. Are there any challenges associated with "W.8.8.19 Integration of 2 to \infty"?

Yes, there are several challenges associated with solving W.8.8.19 Integration of 2 to \infty. These include understanding and applying the appropriate techniques, dealing with infinite values, and determining if the integral converges or diverges. It is essential to have a strong understanding of calculus and advanced mathematical concepts to overcome these challenges.

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