# Finding $\int_{1/4}^4 \frac{1}{x}\sin(x-\frac{1}{x})\,dx$

Question:

Find the value of: $$\int_{1/4}^4 \frac{1}{x}\sin(x-\frac{1}{x})\,dx$$

Attempt: I seriously have no idea how to attempt this question. I suppose I will first have to treat this as an indefinite integral to find the integration of the function, and then substitute the limits to find the value. I tried to do this using the by-parts method, however to no avail. I can't think of what to substitute in order to solve the integral. Any hint would be appreciated.

• Wolfram suggests that there isn't an elementary antiderivative. It might be worth considering an expansion of some sort of the sine term. Commented Jul 13, 2015 at 15:21
• But the limits suggest something nice, maybe try put $x=\frac{1}{t}$ Commented Jul 13, 2015 at 15:22
• Indeed they do. That's a nice spot. (I was referring to the indefinite integral when I mentioned there not being an elementary antiderivative.) Commented Jul 13, 2015 at 15:24

If we substitute $t=x-\frac{1}{x}$ (we are allowed since $g(x)=x-\frac{1}{x}$ is increasing and differentiable over $\left[\frac{1}{4},4\right]$), we are left with: $$I = \int_{-15/4}^{15/4}\sin(t)\frac{dt}{\sqrt{4+t^2}}$$ hence we are integrating an odd integrable function over a symmetric domain with respect to the origin, so the value of the integral is simply zero.
if $I$ is your given integral, simply substitute $1/x = u$ and you will get $-I$, which means the value of the definite integral $I$ is zero.
• True, but only if you prove that the given function is integrable. Otherwise, the risk is to state something like $$\int_{-1}^{1}\frac{dx}{x}=0$$ that is not exactly true if $\int$ stands for the usual Riemann integral. Commented Jul 13, 2015 at 15:28