comparing two integrals: a cancellation method Problem: whether proving $\int x\,f(x)\,\mathrm{d}x &lt; \int x\,g(x)\,\mathrm{d}x$ is equivalent to proving $\int \, f(x)\,\mathrm{d}x &lt; \int g(x)\,\mathrm{d}x$?
I have tried this and want to make sure this is right
Since the range is from $-\infty$ to $\infty$, the answer is no. Here is a counter example. Consider $f(x) = \frac{1}{1+x^4}$ and $g(x) = \frac{x}{1+x^4}$. Then
$$
\int_{-\infty}^{\infty}xf(x)dx = \int_{-\infty}^{\infty}\frac{x}{1+x^4}dx = 0 &lt; \int_{-\infty}^{\infty}xg(x)dx = \int_{-\infty}^{\infty}\frac{x^2}{1+x^4}dx = \frac{\pi}{\sqrt{2}}
$$
And yet
$$
\int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^{\infty}\frac{1}{1+x^4}dx = \frac{\pi}{2} &gt; \int_{-\infty}^{\infty}g(x)dx = \int_{-\infty}^{\infty}\frac{x}{1+x^4}dx = 0
$$
NOTE: I will not go through the steps to get these integrals. They are ugly. The general idea is just to have $f(x)$ be an even function and $g(x)$ be an odd one. Then, $xf(x)$ is odd and $xg(x)$ is even. The integral of an odd function (given this range) is 0.
Since the range is from $-\infty$ to $\infty$, the answer is no. Here is a counter example. Consider $f(x) = \frac{1}{1+x^4}$ and $g(x) = \frac{x}{1+x^4}$. Then
$$
\int_{-\infty}^{\infty}xf(x)dx = \int_{-\infty}^{\infty}\frac{x}{1+x^4}dx = 0 &lt; \int_{-\infty}^{\infty}xg(x)dx = \int_{-\infty}^{\infty}\frac{x^2}{1+x^4}dx = \frac{\pi}{\sqrt{2}}
$$
And yet
$$
\int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^{\infty}\frac{1}{1+x^4}dx = \frac{\pi}{2} &gt; \int_{-\infty}^{\infty}g(x)dx = \int_{-\infty}^{\infty}\frac{x}{1+x^4}dx = 0
$$
NOTE: I will not go through the steps to get these integrals. They are ugly. The general idea is just to have $f(x)$ be an even function and $g(x)$ be an odd one. Then, $xf(x)$ is odd and $xg(x)$ is even. The integral of an odd function (given this range) is 0.