Improper Integration: Divergence of x*sin(kx), k is Real | Math Homework Help

[SSKR]

I am in doubt about convergence of the folowing improper integration:
+infinite
/
\ x * sin(kx) dx; (k is real)
/
0

sorry about my sign of integration! I promisse that I will learn how to post correctly!

I used the intuition:
sin is periodic and bounded and x is unbounded. In my intuition this integration diverges. Bur I need rational answer, intuition in mathematics is something of nothing.

 

[Hurkyl]

What is sen?

Have you tried any theorems that let you prove an integral diverges?

 

[LeonhardEuler]

All you need to do to prove that the integral diverges is appeal to the definition of the improper integral:
$$\int_{0}^{\infty}x\sin{x}dx \equiv\lim_{b\rightarrow\infty}\int_{0}^{b}x\sin{x}dx$$
You can evaluate the integral in terms of b using integration by parts and then show that the limit as b approaches infinity does not exist.

 

[Hurkyl]

[But] I need rational answer, intuition in mathematics is something of nothing.

Intuition is good; but you're right, you need a proof in the end.

However, many definitions and theorems in mathematics are designed to capture our intuition. In particular,

sin is periodic and bounded and x is unbounded. In my intuition this integration diverges.

This is good intuition, and I'm sure that there's a theorem that says almost exactly this, but I can't remember what it is off hand.


But, if you can't find the theorem, you could always try to prove it:


Theorem Let $f(x)$ be a nonzero periodic function with period $L$, and let $g(x)$ be an unbounded function. Then, the integral

$$\int_{0}^{+\infty} f(x) g(x) \, dx$$

diverges.


Actually, the theorem I just stated is false. For example, you could let $f(x)$ be the function:

$$f(x) := \left\{ \begin{array}{l l} 0 \quad & \lfloor x \rfloor \mbox{\ is even} \\ 1 \quad & \lfloor x \rfloor \mbox{\ is odd} \end{array}$$

And let $g(x)$ be the function $g(x) := f(x - 1) \cdot x$


But a suitable modification to the "theorem" I stated (e.g. it would suffice to make $g(x)$ a strictly positive function) would be true. See if you can prove it!

(P.S. what level math are you in?)


LeonhardEuler's approach would work too, so you could do the problem that way. However, it would be much nicer to have a theorem you could use for problems of this type!

(P.S. if you click on any of our math formulae, you will see what you need to type to make that formula)

 

[SSKR]

Thank you for your great help!
I am just a beginner and very lazy!
I think that Lazy is mortal sin for everyone but in special for a mathematician.

The TEX used in this forum is the standart?

I plan to start to learn TEX and this site will help me so much in my learning!

Thank you so much!

 

[SGT]

All you need to do to prove that the integral diverges is appeal to the definition of the improper integral:
$$\int_{0}^{\infty}x\sin{x}dx \equiv\lim_{b\rightarrow\infty}\int_{0}^{b}x\sin{x}dx$$
You can evaluate the integral in terms of b using integration by parts and then show that the limit as b approaches infinity does not exist.

Unfortunately, this integral has no analytical solution. He can try to develop sin x in series, integrate the series and see if it converges.

 

[Tide]

Unfortunately, this integral has no analytical solution. He can try to develop sin x in series, integrate the series and see if it converges.

Yes it does. The integral evaluates to $\sin b - b \cos b$.

 

[HallsofIvy]

Unfortunately, this integral has no analytical solution. He can try to develop sin x in series, integrate the series and see if it converges.

Of course it does: integrate by parts.