Doubt about integral bounds

[dunkelheit]

While finding the antiderivative of

[MATH]\int \frac{\sin x - x \cos x}{x^2+ \sin^2 x} \text{d}x[/MATH]​

I've tried the substitution [MATH]\frac{x}{\sin x}=u[/MATH], so [MATH]\frac{\sin x - x \cos x}{\sin^2 x} \text{d}x=\text{d}u[/MATH]; hence

[MATH]\int \frac{\sin x - x \cos x}{x^2+ \sin^2 x} \text{d}x=\int \frac{\sin x - x \cos x}{\left(\frac{x^2}{\sin^2 x}+ 1\right)\sin^2 x} \text{d}x=\int \frac{1}{u^2+1} \text{d}u=\arctan u +c=\arctan \frac{x}{\sin x} +c[/MATH]​

But if I try to evaluate

[MATH]\int_0^{\infty} \frac{\sin x - x \cos x}{x^2+ \sin^2 x} \text{d}x[/MATH]​

I have some troubles with the integration interval, because when [MATH]x\to 0^+[/MATH] the lower bound in [MATH]u[/MATH] is [MATH]1[/MATH] but

[MATH]\lim_{x \to \infty} \frac{x}{\sin x }=\nexists[/MATH]​

So I don't know how to deal with the upper bound. I suspect that I can't do that substitution in the interval [MATH](0,\infty)[/MATH]; but why? Thanks.

 

[Steven G]

Your substitution was fine. Nice job in finding it.

I am shocked at the trouble that arrives with the change of limits. I do not ever remember seeing this before.

I suggest that you convert back to x's (actually you did just that) and then apply the x-limits. After seeing the answer then maybe we will see what exactly is going on here.

 

[Cubist]

See the graph below. The red line is [math] \arctan(\frac{x}{\sin x})[/math] and the green line is a numerical integral of the original function (with a constant of integration that aligns them at x=0+).



The substitution [math] u=\frac{x}{\sin x} [/math] creates lots of discontinuities. It does seem to re-align after every other discontinuity, but it isn't an ideal result.

I think I found a better substitution to use (a lucky guess based on your choice), try:- [math] u=\frac{\sin x}{x} [/math]

 

[dunkelheit]

@Cubist: Thanks for your answer! Yes, that substitution works and avoid the problem.
I thought about using [MATH]\arctan t + \arctan \frac{1}{t} = \frac{\pi}{2}[/MATH] but it only works for [MATH]t>0[/MATH] and [MATH]\frac{\sin x}{x}[/MATH] change sign infinitely times on [MATH](0,\infty)[/MATH] so I think that it doesn't work. Thanks again!

 

[Cubist]

To show the result of the new substitution...

[MATH]\frac{\sin x}{x}=u[/MATH], so [MATH] \frac{x \cos\left(x\right)-\sin\left(x\right)}{x^2}\text{d}x=\text{d}u[/MATH], hence

[MATH]\int \frac{\sin x - x \cos x}{x^2+ \sin^2 x} \text{d}x[/math]
[math]=-\int \frac{x \cos\left(x\right) - \sin\left(x\right)}{x^2 \left(1+\left(\frac{\sin\left(x\right)}{x}\right)^2\right)} \text{d}x[/math]
[math]=-\int \frac{1}{u^2+1}\text{d}u[/math]
[math]=-\arctan u +c[/MATH]
[math]=-\arctan \left(\frac{\sin x}{x}\right) +c[/MATH]