Question

Interesting advanced calculus question [\(\color{green}{\text{SOLVED}}\)]

Answer 1

\[\large{\color{red}{\text{QUESTION}}}\]

For b > 0, prove that:

1. \(\large{\cfrac{\sin x}{x}dx = \int_{0}^{\infty}(\int_{0}^{b}e^{-ux}\sin x \ dx) \ du}\)

2. \(\large{|\int_{0}^{b}\cfrac{\sin x}{x}dx - \cfrac{\pi}{2}| < \cfrac{1}{b}}\)

Answer 2

\[\large{\color{red}{\text{HINT}}}\]

For y > 0 prove that:

\[\large{\int_{0}^{y}(\int_{0}^{b}e^{-ux}\sin x \ dx) \ du = \int_{0}^{b} (1-e^{-xy})\cfrac{\sin x}{x}dx}\]

and then show that:

\[\large{\lim_{y \rightarrow \infty}\int_{0}^{b} (1-e^{-xy})\cfrac{\sin x}{x}dx = \int_{0}^{b} \cfrac{\sin x}{x}dx}\]

Answer 3

\[\large{\color{red}{\text{SOLUTION}}}\]

First of all we will prove that for b > 0

\[\large{\lim_{y \rightarrow \infty}\int_{0}^{b} (1-e^{-xy})\cfrac{\sin x}{x}dx = \int_{0}^{b} \cfrac{\sin x}{x}dx}\tag{1}\]

We have for y > 0:

\[\large{|\int_{0}^{b}\cfrac{\sin x}{x}dx - \int_{0}^{b}(1-e^{-xy})\cfrac{\sin x}{x}dx|}\]

\[\large{=|\int_{0}^{b}e^{-xy}\cfrac{\sin x}{x}dx|}\]

\[\large{\le \text{max}_{0 \le x \le b} |\cfrac{\sin x}{x}|\int_{0}^{b}e^{-xy} dx}\]

\[\large{= M(b) \ [\cfrac{e^{-xy}}{-y}]_{0}^{b}}\]

\[\large{= M(b) \ \cfrac{(1-e^{-by})}{y}}\]

\[\large{\le \cfrac{M(b)}{y}}\]

Letting \(\large{y \rightarrow \infty}\) we obtain equation 1... \(\large{\color{red}{;)}}\)

Next we have:

\[\large{\int_{0}^{b} (1-e^{-xy})\cfrac{\sin x}{x} dx}\]

\[\large{=\int_{0}^{b}(\int_{0}^{y} xe^{-xu}du)\cfrac{\sin x}{x} dx}\]

\[\large{=\int_{0}^{y}(\int_{0}^{b}e^{-ux}\sin x dx) du}\]

Again letting \(\large{y \rightarrow \infty}\) we obtain:

\[\large{\int_{0}^{b}\cfrac{\sin x}{x} dx = \int_{0}^{\infty}(\int_{0}^{b}e^{-ux}\sin x dx) du}\]

\[\large{=\int_{0}^{\infty} \cfrac{(1-e^{-bu}(u\sin b + \cos b))}{1+u^2} du}\]

\[\large{=\cfrac{\pi}{2} - \int_{0}^{\infty} \cfrac{e^{-bu}(u \sin b + \cos b)}{1+u^2} du}\]

Thus we have:

\[\large{|\int_{0}^{b}\cfrac{\sin x}{x} dx - \cfrac{\pi}{2}|}\]

\[\large{=|\int_{0}^{\infty}\cfrac{e^{-bu}(u\sin b + \cos b)}{1+u^2} du|}\]

\[\large{\le \int_{0}^{\infty}e^{-bu} \cfrac{\sqrt{1+u^2}}{1+u^2} \ du}\]

\[\large{\le \int_{0}^{\infty}e^{-bu} du}\]

\[\large{= \cfrac{1}{b}}\]

\[\large{\color{red}{\text{Hence Proved}}}\]

\[\large{\color{blue}{\text{- Prepared using Notepad++}}}\]

Answer 4

@Kainui @ganeshie8 Any other ideas/suggestion or anything :) ?

Answer 5

@ikram002p

Answer 6

What is this?

Answer 7

This is...uhm... @ganeshie8 what is this ?

Answer 8

This is \(\large{\color{red}{\text{MESS !!!!}}}\) @ℐℵK℣ⱱøƴȡ◆

Answer 9

Right =P

Answer 10

:D

Answer 11

I think I might have something for you give me a second! =)

Answer 12

Sure:)

Answer 13

i smell series/complex stuff from Kainui xD

Answer 14

He is making me nervous :( :P

Answer 15

And now he has stopped typing ... And now I am damn nervous :P :'(

Answer 16

Wait I messed up sorry guys to get you all riled up for nothing... X_X

AHHHH!!! Sorry!!!

Answer 17

\[\color{red}{\text{ITS OKAYYYYYYYYYYY}}\] ;)

Answer 18

But what were you going to type @Kainui O.o

Answer 19

One second, I might have found my error.

Answer 20

kk

Answer 21

i loved this question :)

Answer 22

No, what I was thinking in my mind is exactly what you have done. I had thought there was another way that was similar which looked like this, but once I got closer I realized that I had remembered it incorrectly and your way was how I had seen it before. This is what I had typed:

--

\[\Large I(t)=\int\limits_0^\infty \frac{\sin(tx)}{x}dx\] So we see that our problem is: \[\Large I(1)=\int\limits_0^\infty \frac{\sin(x)}{x}dx\]
We differentiate I(t) with respect to t under the integral sign:
\[\Large I'(t)=\int\limits_0^\infty \cos(tx)dx\]
\[\Large I'(t)= \frac{1}{t}\sin(\infty)\] This is the same as \[\Large -1 \le I'(t) \le 1\] or \[\Large | \frac{1}{t} | \le 1\]

Integrating with respect to t we have:

\[\Large |\ln t +C | \le 1\]

---

And here I stopped because something seemed wrong. X_X

Answer 23

Thanks @ikram002p :)

@Kainui What went wrong ?

Answer 24

this is remineds me with fourier somehow ,,there is something like this mm
(trying to remember )

Answer 25

I guess I could continue with this and get to a result worth something:

\[\Large I(1)=\ln (1) + C = \int\limits_0^\infty \frac{\sin x}{x}dx\] \[\Large |\int\limits_0^\infty \frac{\sin x}{x}dx | \le 1\]

I can't really see how to get anywhere further than this.

Answer 26

mmm can we use limits xd

Answer 27

Well we were using limits @ikram002p :P

Answer 28

hehe no other way than u did xD

Answer 29

:D LOL you would have seen that in my previous question ;)

Answer 30

well no im talking about something else , ill write down in my notes ,and ill post if it went correct :)

Answer 31

Good idea :)

Answer 32

hehe what i did is convert to comlex
integral of e^i z /z
but i couldnt rid of i even after finding residues when =0 mmm xD

Answer 33

Yeah I think I did something similar to you @ikram002p and it didn't work out haha. I separated it out and tried to make it into partial fractions and they ended up having complex coefficients. I guess that doesn't work.

Answer 34

Nothing wrong in trying though :) @ikram002p

I am looking for some more possible methods as that will help me in having alternatives. Having another plan is always a good idea :D

Answer 35

yep , @Kainui i think its cuz complex integral help when we have a,b for integral mm so we could rid of complex coefficients

Answer 36

That would require some more work and then some more and more and some more ;)

Answer 37

I have found some interesting things with power series just now while looking at this. I will continue to see what I can find and maybe we'll have something fun to look at. =)

Answer 38

Great!!!!

Answer 39

Your integrals look a little weird latex-wise, you have the upper and lower sums not on top and below the integral sign.
I would suggest for a number `1.` or `2.` (looking at your first reply) to write the integrals your are having in an equation editor , and in front of the integral type `1.` and after it type `~` for space, you can type `~` more times for more space.

On this site you won't get nice looking integrals or sigmas without an equation editor.

Answer 40

Thanks a lot @SolomonZelman . I was trying to figure out a way to do it since a long time.

Thanks

Answer 41

Anytime ... you can add before your integrals in this case \color{rsvssd}{ `1.~~~ ` } (I grayed the text you type
And you can see it is without a specific color. It will be the closest to the numbers that look like the ones that are without the equation editor.
No color, or some meaningless letters, or when you write " s `green` " (green but messed up (with an `s` or other not needed letter) in the color code space will give a gray-ish color

Answer 42

or you can type them without an equation editor.

Answer 43

The integrals I mean

Answer 44

Okay I will try to use them :)

Answer 45

You can do `\(\large\color{blue}{ ^{ upper sum } }\)` and `\(\large\color{blue}{ _{ lower sum } }\)` but the sums will be a bit shifted though.

Answer 46

Whatever you prefer, good luck and yw :)

Answer 47

:)