Why is $\int_0^\infty(\frac{1}{1+x^a}-\frac{1}{1+x^b})\frac{dx}{x}=0$ for all a,b>0?

Question

I found an equation similar to the one in the question in a table of integrals, and WolframAlpha confirms that the equation holds for all values I've tried, but I haven't been able to prove it.

I tried using the substitution $u=x^a,\frac{du}{au}=\frac{dx}{x}$ on the divergent integral $$\int_0^\infty\frac{dx}{(1+x^a)x},$$ which gives $$\int_0^\infty\frac{du}{(1+u)au},$$ but this implies that $$\int_0^\infty \left(\frac{1}{1+x^a}-\frac{1}{1+x^b}\right)\frac{dx}{x}=\left(\frac{1}{a}-\frac{1}{b}\right)\int_0^\infty\frac{du}{(1+u)u}$$ for all a,b, which is not in general $0$.

Answer 1

In fact $$\begin{eqnarray} I&:=&\int_0^\infty(\frac{1}{1+x^a}-\frac{1}{1+x^b})\frac{dx}{x}\\\ &=&\int_0^\infty\frac{x^{b-1}-x^{a-1}}{(1+x^a)(1+x^b)}dx\\ &\overset{x\to\frac1x}=&\int_0^\infty\frac{x^{-b+1}-x^{-a+1}}{(1+x^{-a})(1+x^{-b})}\frac{dx}{x^2}\\ &=&\int_0^\infty\frac{x^{a+b}(x^{-b-1}-x^{-a-1})}{(1+x^a)(1+x^b)}dx\\ &=&\int_0^\infty\frac{x^{a-1}-x^{b-1}}{(1+x^a)(1+x^b)}dx\\ &=&-I \end{eqnarray}$$ and hence $I=0$.

Answer 2

Partial Fractions $$ \begin{aligned} \int_0^{\infty}\left(\frac{1}{1+x^a}-\frac{1}{1+x^b}\right) \frac{d x}{x} = & \int_0^{\infty}\left[\left(\frac{1}{x}-\frac{x^{a-1}}{1+x^a}\right)-\left(\frac{1}{x}-\frac{x^{b-1}}{1+x^b}\right)\right] d x \\ = & \int_0^{\infty}\left(\frac{x^{b-1}}{1+x^b}-\frac{x^{a-1}}{1+x^a}\right) d x \\ = & {\left[\frac{1}{b} \ln \left(1+x^b\right)-\frac{1}{a} \ln \left(1+x^a\right)\right]_0^{\infty} } \\ = & {\left[\ln \frac{\left(1+x^b\right)^{\frac{1}{b}}}{\left(1+x^a\right)^{\frac{1}{a}}}\right]_0^{\infty} } \\ = & 0 \end{aligned} $$

Answer 3

Your reasoning is flawed because, as you noticed yourself, $\int_0^\infty\frac{du}{(1+u)u}$ is divergent.

Define, for $a>0$, $$F(a):=\int_0^\infty\left(\frac1{1+x^a}-\frac1{1+x}\right)\frac{dx}x.$$ Then, $$\begin{align}F'(a)&=\int_0^\infty\frac{-\ln x}{(x^{a/2}+x^{-a/2})^2}\frac{dx}x\\ &=\int_{-\infty}^\infty\frac{-u}{(e^{au/2}+e^{-au/2})^2}du\\&=0, \end{align}$$ as the integral over $\Bbb R$ of an odd function.

Edit. However, @MartinR's idea is more direct: $$\begin{align}F(a)&=\int_{-\infty}^\infty\frac{e^{(1-a)u/2}-e^{-(1-a)u/2}}{e^{(a+1)u/2}+e^{-(a+1)u/2}+e^{(a-1)u/2}+e^{-(a-1)u/2}}du\\&=0, \end{align}$$ as the integral over $\Bbb R$ of an odd function.

Answer 4

Substituting $x=e^u$ gives $$ I = \int_0^\infty \left(\frac{1}{1+x^a}-\frac{1}{1+x^b} \right)\frac{dx}{x}= \int_{-\infty}^\infty f(u) \, du $$ where $$ \begin{align} f(u) &= \frac{1}{1+e^{au}} - \frac{1}{1+e^{bu}} \\ &= \left( \frac{1}{1+e^{au}} - \frac 12 \right) - \left( \frac{1}{1+e^{bu}} - \frac 12 \right) \\ &= -\frac 12 \tanh\left(\frac{au}{2}\right) + \frac 12 \tanh\left(\frac{bu}{2}\right) \end{align} $$ is an odd function, so that $\int_{-\infty}^\infty f(u) \, du = 0$.

Answer 5

Rewrite the integral as $$- \int_0^\infty \int_a^b \frac{x^{t - 1} \log x}{(1 + x^t)^2} \,dt \,dx = - \int_a^b \int_0^\infty \frac{x^{t - 1} \log x}{(1 + x^t)^2} \,dx \,dt \stackrel{e^{2 u} := x^t}= -\int_a^b \frac1{t^2} \int_{-\infty}^\infty u \operatorname{sech}^2 u \,du \,dt .$$ But the domain of integration of $\int_{-\infty}^\infty u \operatorname{sech}^2 u \,du$ is symmetric about $0$, and its integrand is (integrable and) odd, hence its value is $0$.

Answer 6

Another approach is to use the Mellin transform $$\int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \, \mathrm dx = \frac{\pi}{\nu} \, \csc \left(\frac{\pi \mu}{\nu} \right), \quad 0 < \mu < \nu. $$

See this question for a few proofs.

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Assume that $0 < s \le \delta <\operatorname{min}(a,b)$.

Then $$ \begin{align} \int_{0}^{\infty} \left(\frac{1}{1+x^{a}}- \frac{1}{1+x^{b}} \right)\frac{\mathrm dx}{x} &\overset{\clubsuit}{=} \lim_{s \to 0^{+}}\int_{0}^{\infty} \left(\frac{1}{1+x^{a}}- \frac{1}{1+x^{b}} \right) x^{s-1} \, \mathrm dx \\ &= \lim_{s \to 0^{+}} \left( \int_{0}^{\infty} \frac{x^{s-1}}{1+x^{a}} \, \mathrm dx - \int_{0}^{\infty}\frac{x^{s-1}}{1+x^{b}} \, \mathrm dx \right) \\ &= \lim_{s \to 0^{+}} \left(\frac{\pi}{a} \, \csc \left(\frac{\pi s}{a} \right) - \frac{\pi}{b} \, \csc \left(\frac{\pi s}{b} \right) \right) \\ &\overset{\spadesuit}{=} \lim_{s \to 0^{+}} \left(\frac{\pi}{a} \left(\frac{a}{\pi s} + O(s) \right) - \frac{\pi}{b} \left(\frac{b}{\pi s} + O(s) \right)\right) \\ &= \lim_{s \to 0^{+}} \left(0 + O(s) \right) \\ &= 0. \end{align}$$

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$\clubsuit$ On the interval $(0, 1]$, $\left(\frac{1}{1+x^{a}}- \frac{1}{1+x^{b}} \right) x^{s-1} $ is dominated by the integrable function $\left|\frac{1}{1+x^{a}}- \frac{1}{1+x^{b}} \right| \frac{1}{x}$. And on the interval $(1, \infty)$, $\left(\frac{1}{1+x^{a}}- \frac{1}{1+x^{b}} \right) x^{s-1}$ is dominated by the integrable function $\left|\frac{1}{1+x^{a}}- \frac{1}{1+x^{b}} \right| x^{\delta-1} $.

$\spadesuit$ https://mathworld.wolfram.com/Cosecant.html (8)