Question

improper intergral

Answer 1

\[\huge \text {prove that }\huge \color{green}{ \int\limits _0^{\infty}\frac{ \sin x }{ x }dx} \text { converges}\]

Answer 2

1 fisrt step i did
\[\int\limits_0^{\infty}\left| \frac{\sin x}{x} \right|\]

Answer 3

\[\le \int _0^{\infty} \frac{1}{\left| x \right|}\]

Answer 4

but this is not correct is the enything wrong with this assumption

Answer 5

yeah that won't work because that integral does not converge
i think you need to expand in a power series, then integrate term by term

Answer 6

i see in the book something like cahnging the intergral to sums

Answer 7

don't use the absolute value, if you expand in a power series the terms alternate

Answer 8

yes, "sums" is the power series for \(\frac{\sin(x)}{x}\)

Answer 9

take the power series expansion for sine, then divide everything by \(x\)

Answer 10

\[\sum_{0}^{\infty}\int\limits _{n \pi}^{\pi(n+1)}\left| \frac{ \sin x }{ x } \right|dx\]

Answer 11

i didnt do power series so kinda confused

Answer 12

oh okay
then i guess you are to break it up as you have above

Answer 13

but how do we do that,break up

Answer 14

when you wrote it as a sum, you already broke it up

Answer 15

actually my question is why those boundaries

Answer 16

\[ n\pi ,(n+1)\pi\]

Answer 17

oh, because that is the gimmick that is going to work, that is all. it is a trick
you can bound each of those integrals, then get a bound for the sum

you are not going to find a closed form for \(\int \frac{\sin(x)}{x}dx\) so forget that method
you are going to find a bound for it on each of those intervals

Answer 18

i guess the motivation for those intervals is that you actually know how to evaluate sine at these points

Answer 19

ie 0 and pi

Answer 20

for example, consider
\[\int_{\pi}^{2\pi}\frac{\sin(x)}{x}dx\]

Answer 21

the largest this can be is \(\pi\times \frac{1}{\pi}=1\)

Answer 22

yes thats works the intervals are correcly chosen
how do i convert that intergral to boundswith 0 and pi

Answer 23

now you are making me think

Answer 24

substitution

Answer 25

man power series would really be a help for this one

Answer 26

okay i would try to learn the series thanks and hopefully it will work

Answer 27

that integral is of course improper, but not a problem since
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

Answer 28

look at the series for \(\sin(x)\) then divide it by \(x\) and you will see that the first integral converges
in fact, it is \(\frac{\pi}{2}\) a well known but not obvious result