Is $\int_{-y}^y \frac{v}{x+v}dv < 0$, for $x>1$ and $y>0$?

Question

I got the expression $2y + x\ln\left(\frac{x-y}{x+y}\right) < 0$ after solving, but I don't know how to continue. I tried different random values for $y$ and $x$ to find a contradiction, but the inequality still holds. Any ideas or help to continue the proof/contradiction would be appreciated!

Answer 1

I assume another condition is that $x \gt y$, since this is required for the integral to be well-defined (i.e., the denominator of the fraction in the integration doesn't become $0$, and the argument to $\ln$ in the result you got is $\gt 0$). If so, then rather than explicitly determining the integral result, by splitting the integral and using a substitution of $w = -v \;\;\to\;\; dw = -dv$, we instead get

$$\begin{equation}\begin{aligned} f(x,y) & = \int_{-y}^{y}\left(\frac{v}{x+v}\right)dv \\ & = \int_{-y}^{0}\frac{vdv}{x+v} + \int_{0}^{y}\frac{vdv}{x+v} \\ & = \int_{y}^{0}\frac{(-w)(-dw)}{x-w} + \int_{0}^{y}\frac{vdv}{x+v} \\ & = \int_{0}^{y}\frac{-vdv}{x-v} + \int_{0}^{y}\frac{vdv}{x+v} \\ & = \int_{0}^{y}\left(\frac{-v}{x-v} + \frac{v}{x+v}\right)dv \\ & = \int_{0}^{y}\left(\frac{-2v^2}{(x-v)(x+v)}\right)dv \end{aligned}\end{equation}$$

Since $\frac{-2v^2}{(x-v)(x+v)} \lt 0$ for all $v \in (0, y]$, then $f(x,y) \lt 0$.

Answer 2

We have that for $x>y$

$$\int_{-y}^y \frac{v}{x+v}dv = \left[v-x\log (x+v)\right]_{-y}^y=2y-x\log{\frac{x+y}{x-y}}=y\left(2-\frac xy\log{\frac{\frac x y+1}{\frac x y-1}}\right)$$

and by $u=\frac x y>1$ we consider

$$f(u)=2-u\log{\left(\frac{u+1}{u-1}\right)}=2-\log \left(1+\frac{2}{u-1}\right)^u< 0$$

indeed for $u>1$

$$\left(1+\frac{2}{u-1}\right)^u > e^2 \iff 1+\frac{\frac 2 u}{1-\frac 1u}> e^{\frac 2u}$$

as proved for example here in a fine way by Christian Blatter

• prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$