MHB W.8.8.19 Integration of 2 to \infty

[karush]

$\tiny{w.8.8.19}$
$\textsf{Integration of $2$ to $\infty$}$
$$\displaystyle
I_{19} = \int_{2}^{\infty} \frac{\cos\left({\pi/x}\right)}{{x}^{2}}\,dx $$

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

 

[MarkFL]

$\tiny{w.8.8.19}$
$\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:

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

Next, I would use the substitution:

$$u=\frac{1}{x}\implies du=-\frac{1}{x^2}\,dx$$

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

 

[karush]

I would begin by writing:

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

Next, I would use the substitution:

$$u=\frac{1}{x}\implies du=-\frac{1}{x^2}\,dx$$

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 $u$

 

[MarkFL]

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

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:

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

What do we get when we integrate within the limit?

 

[karush]

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

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

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?

 

[MarkFL]

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:

$$I=\lim_{t\to0}\left(\int_t^{\frac{1}{2}} \cos(\pi u)\,du\right)=\lim_{t\to0}\left(\frac{1}{\pi}\int_t^{\frac{1}{2}} \cos(\pi u)\,\pi\cdot du\right)=\lim_{t\to0}\left(\frac{1}{\pi}\left[\sin(\pi u)\right]_t^{\frac{1}{2}}\right)=\lim_{t\to0}\left(\frac{1}{\pi}\left(\sin\left(\frac{\pi}{2}\right)-\sin(t)\right)\right)=\lim_{t\to0}\left(\frac{1}{\pi}\left(1-\sin(t)\right)\right)=\frac{1}{\pi}$$

 

[karush]

is this known as a Improper Integral?

 

[MarkFL]

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