Interesting problematic integral

[Nebuchadnezza]

This is not a bookquestion by far. It does have a nice solution, but I am stumpled on how to get it

$$\int_{0}^{\infty} 1 - x \, \sin \left( \frac{1}{x} \right) \, \text{dx}$$

Now I could split the integrals into two pieces, where the first part obviously goes towards infinity. The second integral also seem to tend to infinity but at a slower pace, which makes the integral converge.

I tried to make a serie expansion of the sine, but that did not work out. The answer is supposed to be $$\frac{\pi}{4}$$

Any hints, tips or solutions?

Another problem is that $$x \, \sin \left( \frac{1}{x} \right)$$ oscillates faster and faster when approaching zero. The limit is zero, but this confuses me...

 

[Eynstone]

Is the integrand (1-x)sin(1/x) or 1 - xsin(1/x) ? Please clarify.

 

[micromass]

Hmm, this integral reminds me of the integral

$$\int_0^{+\infty}{\frac{\sin(x)}{x}dx}=\frac{\pi}{2}$$

This integral is well-documented and if you want, I can provide you a link on how to find it.

So my suggestion is to do something similar. Maybe your first step should be the substitution $$u=1/x$$...

 

[micromass]

The following wiki can be interesting: http://en.wikipedia.org/wiki/Dirichlet_integral

 

[JJacquelin]

It is true that the value of the integral is pi/4. (Proof in attachement)

 

[Nebuchadnezza]

a micromass, the integral is the same as JJacquelin solved. I am familiar with that problem, and I have solved that problem earlier. See this thread

https://www.physicsforums.com/showthread.php?t=490052

Now nice solution JJackuelin, but my question is. Is it possible to evaluate this integral without the use of special functions ?

I think I made some progress atleast...

I can rewrite $1-x\sin\left(\frac1x\right)$ as $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}\left(\frac1x\right)^{2k}}{(2k+1)!}$

Then my integral turns into

$$I = \left[\sum_{k=1}^{\infty}\frac{(-1)^{k+2}\left(\frac1x\right)^{2k-1}}{(2k+1)!(2k-1)}\right]_0^{\infty}$$

Which obviously is $0$ when $\lim_{x \to \infty}$

Though I have no idea how to evaluate that mess above...