Find the following integral, where $a$ is a real number bigger than $1$: $$\int_1^{a^2} \frac{\ln x}{\sqrt x(x + a)}\,\mathrm dx.$$

By using the substitution $t = \sqrt x$, I got this new integral which seems to be easier to solve, but I haven't found any way to do it yet: $$4\int_1^a \frac{\ln t}{t^2 + a}\,\mathrm dt.$$

Thank you in advance!

  • $\begingroup$ I think you shouldn't have a factor of 4 there. Also, wolframalpha says that the integral involves the polylogarithm function, so it's likely that there is no elmentary antiderivative. $\endgroup$
    – user159517
    Commented Feb 21, 2017 at 12:19

1 Answer 1


Let $t=\sqrt{x}$ \begin{equation} I = \int\limits_{1}^{a^{2}} \frac{\ln x}{\sqrt{x}(x+a)} dx = 4 \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt \end{equation}

Integrating by parts, we have \begin{align} I_{1} &= \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt \\ &= \frac{\ln t}{\sqrt{a}} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) \Big|_{1}^{a} \, - \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt \\ &= \frac{\ln a}{\sqrt{a}} \tan^{-1}(\sqrt{a}) \, - \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt \end{align}

Let $y=t/ \sqrt{a}$ \begin{align} I_{2} &= \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt = \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{1}{y} \tan^{-1}(y) dy \\ \tag{a} &= \frac{i}{2} \left[ \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1-iy)}{y} dy \, - \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1+iy)}{y} dy \right] \\ \tag{b} &= \frac{i}{2} \left[ \mathrm{Li}_{2}(-iy) - \mathrm{Li}_{2}(iy) \right] \Big|_{1/\sqrt{a}}^{\sqrt{a}} \\ &= \frac{i}{2} \left( \left[ \mathrm{Li}_{2}(-i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{i}{\sqrt{a}}\right) \right] - \left[ \mathrm{Li}_{2}(i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{-i}{\sqrt{a}}\right) \right] \right) \\ \tag{c} &= \frac{i}{2} \left( \left[ -\frac{\pi ^{2}}{6} - \frac{1}{2} \ln ^{2}(i\sqrt{a}) \right] - \left[ -\frac{\pi ^{2}}{6} - \frac{1}{2} \ln ^{2}(-i\sqrt{a}) \right] \right) \\ \tag{d} &= \frac{\pi}{4} \ln a \end{align}

a. $\tan^{-1}(y) = \frac{i}{2} [\ln (1-iy) - \ln (1+iy)]$

b. Dilogarithm function \begin{equation} \mathrm{Li}_{2}(z) = -\int_{0}^{z} \frac{\ln (1-x)}{x} dx \end{equation}

c. Use the identity \begin{equation} \mathrm{Li}_{2}(z) + \mathrm{Li}_{2}(1/z) = -\frac{\pi ^{2}}{6} - \frac{1}{2} \ln ^{2}(-z) \end{equation}

d. $\ln (\pm iz) = \ln z \pm i\pi /2$

Now we have \begin{equation} I = 4I_{1} = \frac{4}{\sqrt{a}} (\ln a) \tan^{-1}(\sqrt{a}) \, - \frac{\pi}{\sqrt{a}} \ln a \end{equation}

  • $\begingroup$ Really really nice. (+1) $\endgroup$
    – clathratus
    Commented Nov 30, 2018 at 5:08

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