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If $2f(x)+f(-x)=(1/x)\sin(x-(1/x))$. Then what is the value of: $$\int_{e^{-1}}^{e} f (x) \mathrm {d}x $$

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    First step is to find f(x) – 2017-01-24
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    How do we know its an even or an odd function? – 2017-01-24
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    I think its an even function – 2017-01-24
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    As the LHS does not change on putting x=(-x) – 2017-01-24
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    Considering that f(x) is 1/3 of LHS. Now how do we integrate this integral? – 2017-01-24

2 Answers 2

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$$2f(x)+f(-x)=\left(\frac{1}{x}\right)\sin\left(x-\frac{1}{x}\right)\\\text{Change $x$ to $-x$}\\\begin{align}2f(-x)+f(+x)&=\left(-\frac1x\right)\sin\left(-x+\frac{1}{x}\right)\\&=-\frac1x\cdot\sin\left(-\left(x-\frac1x\right)\right)\\&=\frac1x\cdot\sin\left(x-\frac1x\right)\end{align}$$ Solve this system of equations to find $f(x)$. $$\begin{cases}2f(x)+f(-x)=\dfrac1x\sin\left(x-\dfrac1x\right)\\f(x)+2f(-x)=\dfrac1x\sin\left(x-\dfrac1x\right)\end{cases}\\\to f(x)=\dfrac13\cdot\frac1x\cdot\sin\left(x-\dfrac1x\right)$$ Now, we can solve the integral. $$u=\frac1x \to dx=-\frac{1}{x^2} ,\\x=e=\frac1u \to u=e^{-1} \\x=e^{-1}=\frac1u \to u=e\\ \to \\ A=\int_{e^{-1}}^{e} \dfrac13\cdot\frac1x\cdot\sin\left(x-\dfrac1x\right)=\\ \int_{e}^{e^{-1}} \dfrac13\cdot u\cdot \sin\left(\frac1u-u\right)\left(-\dfrac{1}{u^2}du\right)=\\ \int_{e}^{e^{-1}} \dfrac13\cdot\frac1u\cdot\sin\left(u-\frac1u\right)du=\\ -\int_{e^{-1}}^{e} \dfrac13\cdot\frac1u\cdot\sin\left(u-\frac1u\right)du=\color{red} {-A} \\\to -A=A \to 2A=0\to A=0$$

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    Can u give me a hint to proceed with this integration @Khosrotash .some sort of substitution? – 2017-01-24
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    Take $u=\dfrac{1}{x} \to x=\dfrac{1}{u} \\dx=-\dfrac{1}{u^2}$ then put them in integral , after that you have the same integral by $-1$ coefficient . sum of them is zero . so you answer is $\frac{0}{2}=0$ – 2017-01-24
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  1. we have $f(-x)=f(x)$.

  2. with the substition $t=\frac{1}{x}$ show that $\int_{e^{-1}}^{e} (1/x) \sin(x-(1/x)) dx=0$

  3. Use 1. and 2. to compute $\int_{e^{-1}}^{e} f(x) dx$

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    We have $\int_{e^{-1}}^{e} (1/x) \sin(x-(1/x)) dx=0$ – 2017-01-24