Sine of e?

Calculus Level 3

If $2f(x) + f(-x) = \frac 1x \sin\left( \frac{x^2-1}x \right)$ for $x > 0$ , then what is the value of the integral below?

$\large \int_{\frac 1e}^e f(x) \, dx$

Try my set

The answer is 0.

1 solution

Vishwak Srinivasan
May 14, 2015

$2f(x) + f(-x) = \frac{1}{x} \sin(\frac{x^2 - 1}{x}) . . . . . . . . . (1)$

Substitute $x \rightarrow -x$

$2f(-x) + f(x) = \frac{1}{-x}\sin(\frac{x^2 - 1}{-x}) = \frac{1}{x} \sin(\frac{x^2 - 1}{x}) . . . . . . . . (2) \because \sin(-x) = -sin(x) \text{ and the negative signs get cancelled }$

Using $(\text{(1) \& (2))}$ , we get

$f(x) = \frac{1}{3x} \sin(\frac{x^2 - 1}{x})$

$\displaystyle\int _{\frac{1}{e}}^{e} f(x) dx = \int_{\frac{1}{e}}^{e} \frac{1}{3} \sin(\frac{x^2 - 1}{x}) dx$

Substitute $x = e^t \Rightarrow dx = e^t dt$

The limits get changed to $(-1,1)$ and so does the integral.

The changed integral (say $I$ is:

$I = \displaystyle \int_{-1}^{1} \frac{1}{3e^t} \sin(e^t - \frac{1}{e^t}) e^t dt = \int_{-1}^{1} \frac{1}{3} \sin(e^t -\frac{1}{e^t}) dt$

The integrand is an odd function and hence $\boxed{I = 0}$

I did the same as you to find $f(x)$ , but when calculating the Integral, I used that $\displaystyle \int_{a}^{b} f(x) = \int_{a}^{b} f(a+b-x)$ , thus, we have: $\displaystyle I = \int_{\frac{1}{e}}^{e} \frac{1}{3x}sin\left(\frac{x^2-1}{x}\right) = \int_{\frac{1}{e}}^{e} -\frac{1}{3(e+\frac{1}{e}-x)}sin\left(\frac{x^2-1}{x}\right)=-I$ then, $I=-I \Rightarrow 2I=0 \Rightarrow I=0$

Gabriel Benício - 6 years, 1 month ago

You have forgotten to substitute $e + \frac{1}{e} - x \leftarrow x$ in the $\sin$ part. And how is that $-I$ , by the way.?

Vishwak Srinivasan - 6 years, 1 month ago

No, I haven't... In that case, $\displaystyle sin\left( \frac{x^2-1}{x}\right) = sin\left(x-\frac{1}{x}\right)$ , substitute and you have $sin\left(\frac{1}{e} - e + e-\frac{1}{e}-x+\frac{1}{x}\right) = - sin\left(x-\frac{1}{x}\right)$