Interesting problematic integral

Nebuchadnezza
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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...
 
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Is the integrand (1-x)sin(1/x) or 1 - xsin(1/x) ? Please clarify.
 
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...
 
It is true that the value of the integral is pi/4. (Proof in attachement)
 

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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...
 
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