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How do we solve the following integral ?

$ \int_{0}^{1} 2x\sin(\frac{1}{x}) - \cos(\frac{1}{x})\ dx. $

I tried to proceed by integration by parts but got stuck.

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6 Answers 6

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I think I have a guess for this one. But it is only a guess. This looks like an expansion of product rule. In particular, notice that $$ \frac{d}{dx}\left(x^2\sin(1/x)\right) = 2x\sin(1/x) + x^2\cos(1/x)(-1/x^2) = 2x\sin(1/x) - \cos(1/x) $$ So, we get $$ \int_0^1 2x\sin(1/x) - \cos(1/x) dx = \left[x^2\sin(1/x)\right]_0^1 = \sin(1) $$ I realize this isn't very rigorous. I think we need to consider a limit for the $0$ side $$ \lim_{x\to 0}x^2\sin(1/x) = 0 $$ which follows from by the squeeze theorem.

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$$\int 2x\sin(1/x)dx -\int \cos(1/x)dx$$

In fact, you can only do integration by parts only once, $$\int 2x\sin(1/x)dx=\int \cos(1/x)dx+x^2\sin(1/x)$$

By Letting $$f'(x)=2x \implies f(x)=x^2$$ $$g(x)=\sin(1/x) \implies g'(x)=-x^2\cos(1/x)$$

$$\therefore \int 2x\sin(1/x)dx -\int \cos(1/x)dx=\int \cos(1/x)dx+x^2\sin(1/x) - \int \cos(1/x)dx$$ $$=x^2\sin(1/x)$$

Now, you need only find the definite integral.

$$\sin(1)-\lim_{x \to 0} x^2\sin(1/x)$$ Using the squeeze theorem,

$$-1<\sin(1/x)<1 \implies 0<\lim_{x \to 0} x^2\sin(1/x)<0$$

So, the limit is equal to zero.

$$\int_{0}^{1} 2x\sin(1/x) - \cos(1/x)\ dx =\sin(1)$$

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You can do this using int by parts only once.

First I split it into a sum of integrals: $$\int_0^1 2x\sin(1/x)dx - \int_0^1\cos(1/x)dx$$

Then integrate the first integral only:

I did this with integration by parts $u = sin(1/x)$ and $dv = 2x$. This gives: $$\left[uv-\int vdu\right] - \int cos(1/x)$$ $$\left[x^2\sin(1/x)|_0^1 - \int_0^1-\frac{x^2}{x^2}\cos(1/x)dx\right] - \int_0^1 \cos(1/x) dx$$ $$x^2\sin(1/x)|_0^1 + \left[\int_0^1\cos(1/x)dx - \int_0^1 \cos(1/x) dx\right]$$

Then the $\int cos(1/x)dx$ cancels out. So evaluate $x^2\sin(1/x)]_0^1$. At $0$ use squeeze theorem as @Alec said.

The final answer is $1\sin(1)$

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Before evaluating the definite integral, I will evaluate the indefinite integral. $$~~\int 2x\sin(1/x) - \cos(1/x)\,dx\\ = \int 2x\sin(1/x) - \int \cos(1/x)\,dx$$

The first integral is evaluated as such:

$$\int 2x\sin(1/x) = 2\int x\sin(1/x)$$ $$ u = x, ~du=dx, ~dw=\sin(1/x), ~w = x \sin(1/x)-Ci(1/x)$$ Where $Ci(x)$ is the cosine integral, $Ci(x) = -\int_x^{\infty}\frac{\cos t}{t}dt$ and $\frac{d}{dx}Ci(x) = \frac{\cos(x)}{x} $

This cos integral is probably the first place you got stuck, it is found by integration by parts: $$\int \sin(1/x)~dx \\ a = \sin(1/x),~ da = -\cos(1/x)/x^2~dx,~ db = dx,~ b = x \\ \int a~db = ab - \int b ~da \\ \int \sin(1/x) = x\sin(1/x)- \int \cos(1/x)/x = x\sin(1/x) - Ci(1/x)$$

Anyway, $$\int x\sin(1/x) = x^2 \sin(1/x)-xCi(1/x) - \int x\sin(1/x)-Ci(1/x)~dx$$

Next, By a similar method, $$\int \cos(1/x) = \frac{sin(1/x)}{x^2}$$

Therefore, $$\int 2x\sin(1/x) - \cos(1/x)\,dx = \\x^2 \sin(1/x)-xCi(1/x) - \int x\sin(1/x)-Ci(1/x)~dx - \frac{sin(1/x)}{x^2}$$

And I'll leave you to evaluate the rest.

Here are some referances:

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  • $\begingroup$ And mine is not one of them. It does not mean that it isn't a valid solution. $\endgroup$ Commented Dec 15, 2013 at 8:07
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Set $u'=2x$ and $v=\sin(1/x)$ and apply $\int u' v = u v - \int u v'$. Now use the integration bounds.

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Integration by parts $$ \begin{aligned} \int_0^1 2 x \sin \left(\frac{1}{x}\right) d x = & \int_0^1 \sin \left(\frac{1}{x}\right) d\left(x^2\right) \\ = & {\left[x^2 \sin \left(\frac{1}{x}\right)\right]_0^1-\int_0^1 x^2 \cos \left(\frac{1}{x}\right) \cdot\left(-\frac{1}{x^2}\right) d x } \\ = & \sin 1+\int_0^1 \cos \frac{1}{x} d x \end{aligned} $$ Rearranging yields $$\int_0^1 \left(2 x \sin \frac{1}{x}-\cos \frac{1}{x} \right) dx= \sin 1 $$

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