I am trying to evaluate the integral $$ I_k(x)=\int_1^x \log^k t \frac{\sqrt{t-1}}{t^2} dt $$ with $x$ tending to infinity.

In fact, I wish to have an estimate $$ \sum_{k=0}^\infty \frac{1}{\log^k x} \int_1^x \log^k t \frac{\sqrt{t-1}}{t^2} dt = C+E(x), $$ where $C$ is explicit and $E(x) \to 0$ as $x \to \infty$.

The integral inside may be actually rewritten as $$ I_k(x) = \int_1^x \log^k t \sqrt{1-\frac{1}{t}} \frac{dt}{t^{3/2}} = c_k \int_{1/x^2}^1 \sqrt{1-t^2} \log^k t dt, $$ hence can be possibly attacked via multiple zeta values.

Is that possible?


2 Answers 2


First of all, your series diverges for any $x>1$, because we always have

$$I_k(x)\geq \int_{\sqrt{x}}^x \log^k t\sqrt{t-1}t^{-2}dt\geq \frac{\sqrt{\sqrt{x}-1}}{x}\int_{\sqrt{x}}^{x} \frac{\log^k t}{t} dt=\frac{\sqrt{\sqrt{x}-1}}{x}\frac{(\log^{k+1} x)(1-2^{-k-1})}{k+1}.$$

If you want to get the asymptotics for $I_k(x)$ when $k$ is fixed, then using the substitution $t=1/u$ we get

$$I_k(x)=\int_{1/x}^1 (-1)^k\log^k u \sqrt{(1-u)/u}du=(-1)^k(j_k+R_k(x)),$$


$$j_k=\int_0^1 u^{-1/2}(1-u)^{1/2}\log^k u du$$


$$R_k(x)=\int_0^{1/x} u^{-1/2}(1-u)^{1/2}\log^k u du=O_k\left(\frac{\log^{k+1} x}{x^{3/2}}\right).$$

Now lets compute $j_k$ in terms of the derivatives of Gamma-function. By the beta-integral, we have for any $s>-1$:

$$\int_0^1 u^s(1-u)^{1/2}du=B(s+1,3/2)=\frac{\sqrt{\pi}\Gamma(s+1)}{2 \Gamma(s+5/2)}.$$

Now differentiate the lhs of this equality $k$ times and set $s=-1/2$. Using $(u^s)'=u^s\log u$, we conclude that

$$j_k=\left(\frac{\partial^k}{\partial s^k} \left. \frac{\sqrt{\pi}\Gamma(s+1)}{2 \Gamma(s+5/2)}\right)\right|_{s=-1/2}.$$

  • $\begingroup$ The question arises: is $s=\frac 1 2$ small enough? $\endgroup$
    – user64494
    May 12, 2018 at 14:40
  • $\begingroup$ In fact, the equality holds for $\mathrm{Re}\,s>-1$. So, yes, it is) $\endgroup$ May 12, 2018 at 15:04
  • $\begingroup$ Is there a mistake in the change of variables $t=1/u$ or am I being silly? $\endgroup$
    – toshi
    May 12, 2018 at 15:26
  • $\begingroup$ Numerical calculations do not confirm it: $j_2=.5204777983$, whereas NIntegrate[Log[t]^2*Sqrt[t - 1]/t^2, {t, 1, 50}] performs $4.6862 $. $\endgroup$
    – user64494
    May 12, 2018 at 15:26
  • $\begingroup$ @toshi, yes there is a mistake, but I hope now everything is fine. There should be $-1/2$ instead of $1/2$ $\endgroup$ May 12, 2018 at 16:14

In view of the output of the Mathematica code

Table[Series[Integrate[Log[t]^k*Sqrt[-t^2+1],{t, 1/x^2, 1},Assumptions->x> 1],{x,Infinity,2}],{k,1,3}]

$$O\left(\left(\frac{1}{x}\right)^3\right)+\frac{2 \log (x)+1}{x^2}+\left(-\frac{\pi }{8}-\frac{1}{8} \pi \log (4)\right), $$ $$\frac{1}{48} \left(\pi ^3+6 \pi +3 \pi \log ^2(4)+3 \pi \log (16)\right)+\frac{-4 \log ^2(x)-4 \log (x)-2}{x^2}+O\left(\left(\frac{1}{x}\right)^3\right), $$ $$\frac{1}{32} \left(-12 \pi \zeta (3)-\pi ^3-6 \pi -\pi \log ^3(4)-3 \pi \log ^2(4)-\pi ^3 \log (4)-\pi \log (4096)\right)+\frac{8 \log ^3(x)+12 \log ^2(x)+12 \log (x)+6}{x^2}+O\left(\left(\frac{1}{x}\right)^3\right) $$ I have doubts concerning simple asymptotics for $I_k(x)$.


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