The improper integral evaluation

Question

I need to evaluate the following integral: $$I=\int\limits_{0}^{1}\frac{1}{x}\frac{\frac{q}{p}\frac{x^{-p}-1}{x^{-q}-1}-1% }{x^{-q}-1}dx,$$ where $0<p/2<q\leq p$. It seems not to have a closed-form solution (except for $q=p$ when $I=0$). However, showing that it is finite (as calculations show) would be also great. The integrand converges to $\infty$ as $x\rightarrow 0$, and to $0$ as $x\rightarrow 1$.

Answer 1

Rewriting the integrand $$\frac{ qx^q-px^p+(p-q) x^{p+q}} { p\, \left(x^q-1\right)^2 \,x^{p-q+1}}$$ there is an antiderivative which express in term of the Gaussian hypergeometric function. $$\frac{\left(x^p-1\right) x^{q}}{p\, x^p\left(x^q-1\right)}-\frac{x^{q-p}}{p} \, _2F_1\left(1,\frac{q-p}{q};\frac{2q-p}{q};x^q\right)+$$ $$\frac{\log \left(1-x^q\right)}{q}-\frac{\log \left(1-x^q\right)}{p}$$

Using the bounds and simplifying, the definite integral is just $$\large\color{blue}{I=\frac 1q-\frac{p-q}{p\,q} \,H_{-\frac{p}{q}}}$$ where $H_n$ is the generalized harmonic number.

Remember that Euler integral representation of harmonic numbers is $$H_n=\int_0^1 \frac {1-x^n }{1-x }\,dx$$

Looking at the bounds of $q$ and using asymptotics

• If $q=p \left(\frac{1}{2}+\epsilon \right)$ $$p\,I=\frac{1}{4 \epsilon }+\frac{1}{2}-\frac{2(\pi ^2+6)}{3} \epsilon+4 \left(4 \zeta (3)+\pi ^2+2\right) \epsilon ^2+O\left(\epsilon ^3\right)$$

• If $q=p \left(1-\epsilon \right)$ $$p\,I=\epsilon+\frac{\pi^2+6}6\epsilon^2+\left(\zeta (3)+\frac{\pi ^2}{3}+1\right) \epsilon ^3+O\left(\epsilon ^4\right)$$

Edit

Let $q=\frac 12 p(1+x)$ with $0 < x <1$ and use Wolfram Alpha to plot $$p\,I=f(x)=\frac{2+(x-1) H_{-\frac{2}{x+1}}}{x+1}$$

Just copy/past the next line.

Plot[(2+(x-1)*HarmonicNumber[-2/(1+x)])/(1+x),{x,0,1}]

Answer 2

We know the integrand is continuous on $(0,1]$. So we must investigate the integrand as $x \to 0^+$. The OP says the case $q=p$ is already done. So assume $0 < p/2 < q < p$.

Then as $x \to 0^+$ we have \begin{align} x & \to 0 \\ x^{-p} & \to +\infty\qquad\text{since }-p<0 \\ x^{-p} - 1 & \sim x^{-p} \\ x^{-q} - 1 & \sim x^{-q}\qquad\text{similarly} \\ \frac{x^{-p} - 1}{x^{-q} - 1} & \sim x^{-p+q} \\ \frac{q}{p}\frac{x^{-p} - 1}{x^{-q} - 1} & \sim \frac{q}{p}x^{-p+q} \to +\infty\quad\text{since }-p+q<0 \\ \frac{q}{p}\frac{x^{-p} - 1}{x^{-q} - 1}-1 & \sim \frac{q}{p}x^{-p+q} \\ x^{-q}-1 & \sim x^{-q}\qquad\text{already noted above } \\ \frac{\frac{q}{p}\frac{x^{-p} - 1}{x^{-q} - 1}-1}{x^{-q}-1} &\sim \frac{q}{p} x^{-p+2q} \\ \frac{1}{x}\frac{\frac{q}{p}\frac{x^{-p} - 1}{x^{-q} - 1}-1}{x^{-q}-1} &\sim \frac{q}{p} x^{-p+2q-1} \end{align} But $-p+2q-1 > -1$, so $$ \int_0^1 x^{-p+2q-1}\;dx < + \infty $$ and by comparison, $$ \int_0^1 \frac{1}{x}\frac{\frac{q}{p}\frac{x^{-p} - 1}{x^{-q} - 1}-1}{x^{-q}-1} \;dx < +\infty . $$

Answer 3

Define the harmonic number $H(s)$ by

$$ H(s) = \int_{0}^{1} \frac{1-x^s}{1-x} \, \mathrm{d}x = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+s} \right). $$

When $s = 0, 1, 2, \ldots, $ we easily check that $H(s) = 1 + \frac{1}{2} + \cdots + \frac{1}{s}$. So, the above integral extends the harmonic number for non-integral arguments. Then we prove:

Claim. For $p, q > 0$, OP's integral $I$ converges if and only if $p < 2q$. In this case,

$$ I = \frac{q-p}{qp}\left( H_{1-p/q} - 1 \right). $$

Indeed, let $p, q > 0$. Then with $\alpha = p/q$,

\begin{align*} I &= \int_{0}^{1} \frac{1}{x} \frac{1}{x^{-q}-1}\biggl( \frac{1}{\alpha}\frac{x^{-p}-1}{x^{-q}-1}-1 \biggr) \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{x^{q-1}}{1 - x^q}\biggl( \frac{1}{\alpha}\frac{x^{q-p} - x^q}{1 - x^q} - 1 \biggr) \, \mathrm{d}x \\ &= \frac{1}{q} \int_{0}^{1} \frac{1}{1 - t}\biggl( \frac{1}{\alpha} \frac{t^{1-\alpha} - t}{1 - t} - 1 \biggr) \, \mathrm{d}t \tag{$t = x^q$} \end{align*}

Since the singularity of the integrand at $t = 1$ is removable for any value of $\alpha$, the convergence of $I$ is determined by the singularity of the integrand near $t = 0$. This then shows that the inntegral $I$ converges if and only if $\alpha < 2$, or equivalently, $p < 2q$.

To evaluate the last integral, we consider the corresponding indefinite integral. Perform integration by parts and using $\left( \frac{t}{1-t} \right)' = \frac{1}{(1-t)^2}$, we get

\begin{align*} \int \frac{1}{1 - t}\biggl( \frac{1}{\alpha} \frac{t^{1-\alpha} - t}{1 - t} - 1 \biggr) \, \mathrm{d}t &= \frac{1}{\alpha} \int \frac{t^{1-\alpha} - t}{(1 - t)^2} \, \mathrm{d}t - \int \frac{1}{1 - t} \, \mathrm{d}t \\ &= \frac{1}{\alpha} \frac{t(t^{1-\alpha} - t)}{1 - t} - \int \frac{1}{\alpha} \frac{t((1-\alpha)t^{-\alpha} - 1)}{1 - t} \, \mathrm{d}t - \int \frac{1}{1 - t} \, \mathrm{d}t \\ &= \frac{1}{\alpha} \frac{t^{2-\alpha} - t^2}{1 - t} + \frac{1}{\alpha} \int \frac{(1-\alpha)(1-t^{1-\alpha}) - (1 - t)}{1 - t} \, \mathrm{d}t \end{align*}

So, if $\alpha < 2$, then

\begin{align*} I &= \lim_{\substack{a \to 0^+ \\ b \to 1^-}} \frac{1}{q} \biggl( \biggl[ \frac{1}{\alpha} \frac{t^{2-\alpha} - t^2}{1 - t} \biggr]_{a}^{b} + \frac{1}{\alpha} \int_{a}^{b} \frac{(1-\alpha)(1-t^{1-\alpha}) - (1 - t)}{1 - t} \, \mathrm{d}t \biggr) \\ &= \frac{1}{q} \biggl( 1 + \frac{1-\alpha}{\alpha} H_{1-\alpha} - \frac{1}{\alpha} \biggr) \\ &= \frac{q-p}{qp}\left( H_{1-p/q} - 1 \right) \end{align*}

as required. $\square$

Answer 4

$$I=\int_0^1 x^{2q-p-1}\frac{(p-q)x^p-px^{p-q}+q}{p(1-x^q)^2}dx$$ By $x=u^{1/q}$ substitution $$I=\frac1{pq}\int_0^1\frac{(p-q)u-p+qu^{1-\frac pq}}{(1-u)^2}du$$ After integration by parts $$I=\frac1q-\frac{p-q}{pq}\int_0^1\frac{1-u^{-\frac pq}}{1-u}du\\=\frac1q-\frac{p-q}{pq}H_{-\frac pq}$$ for $p<2q$.