AMM12184

Problem - P. Perfetti (Italy)

Prove

$${\int }_1^{\mathrm{\infty }}\frac{\ln (x^4-2x^2+2)}{x\sqrt{x^2-1}}\mathrm{d}x=\pi \ln (2+\sqrt{2}).$$

Solution

Call the stated integral $I$ and let $x=\sqrt{u+1}$, so $\mathrm{d}x=\frac{\mathrm{d}u}{2\sqrt{u+1}}$ and

$$I=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\ln (u^2+1)}{(u+1)\sqrt{u}}\mathrm{d}u.$$

Now we let $\alpha \ge 0$ and generalize the resulting integral to

$$I(\alpha )=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\ln (\alpha x^2+1)}{(x+1)\sqrt{x}}\mathrm{d}x,$$

so that $I(0)=0$ and $I(1)=I$. Then differentiating under the integral sign gives

$$\begin{array}{r}{I}^{\prime }(\alpha )=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^{3/2}}{(x+1)(\alpha x^2+1)}\mathrm{d}x.\end{array}$$

We now focus on evaluating $I^{\prime }(\alpha )$ by contour integration. To allow for an easier analysis we further restrict $\alpha >0$. Consider the contour integral

$$\begin{array}{}\text{(1)}& \frac{1}{2}{\oint }_{C_{R,ϵ}}\frac{z\cdot z^{1/2}}{(z+1)(\alpha z^2+1)}\mathrm{d}z\end{array}$$

where $R>\text{max}\{1,1/\sqrt{\alpha }\}$, $0<ϵ<\text{min}\{1,1/\sqrt{\alpha }\}$, and $C_{R,ϵ}$ is the following keyhole contour (here the branch cut for $z^{1/2}$ is on the real axis) in the following diagram.

We can evaluate $I^{\prime }(\alpha )$ by calculating $(1)$ in two ways. First, by the residue theorem, the integral $(1)$ is equal to

$$\begin{array}{}\text{(2)}& \underset{\varphi \in \{-1,\frac{1}{\sqrt{\alpha }},-\frac{1}{\sqrt{\alpha }}\}}{2\pi i\sum }\underset{z=\varphi }{\text{Res}}\left(\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\right)=\frac{\pi }{\sqrt{2}}\cdot \frac{{\alpha }^{-3/4}-{\alpha }^{-1/4}}{1+\alpha }+\frac{\pi }{1+\alpha }.\end{array}$$

The second evaluation of $(1)$ can be made by breaking up $C_{R,ϵ}$ as

$${\oint }_{C_{R,ϵ}}={\int }_{L_1}+{\int }_{{\mathrm{\Gamma }}_R}+{\int }_{L_2}+{\int }_{{\gamma }_{ϵ}}$$

and examining how each integral behaves under the limit as $R\to \mathrm{\infty }$ and $ϵ\to 0$. Along the straight path $L_1$ above the real axis we have

$$\begin{array}{rl}\underset{R\to \mathrm{\infty }}{lim}\underset{ϵ\to 0}{lim}{\int }_{L_1}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z& =\underset{R\to \mathrm{\infty }}{lim}\underset{ϵ\to 0}{lim}{\int }_{ϵ}^R\frac{x^{3/2}}{2(x+1)(\alpha x^2+1)}\mathrm{d}x\\ & =I^{\prime }(\alpha ).\end{array}$$

We may bound the integral along ${\mathrm{\Gamma }}_R$ as

$$\begin{array}{rl}\left|{\int }_{{\mathrm{\Gamma }}_R}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z\right|& =\left|\frac{i{R}^{3/2}}{2}{\int }_0^{2\pi }\frac{e^{3i\theta /2}}{(R{e}^{i\theta }+1)(\alpha R^2{e}^{2i\theta }+1)}\mathrm{d}\theta \right|\\ & \le \frac{R^{3/2}}{2}{\int }_0^{2\pi }\left|\frac{1}{(R{e}^{i\theta }+1)(\alpha R^2{e}^{2i\theta }+1)}\right|\mathrm{d}\theta \\ & \le \frac{R^{3/2}}{2}{\int }_0^{2\pi }\left|\frac{1}{(R-1)(\alpha R^2-1)}\right|\mathrm{d}\theta ,\\ & \text{by the reverse triangle inequality}\\ & =\frac{\pi R^{3/2}}{(R-1)(\alpha R^2-1)}.\end{array}$$

So we have

$$\begin{array}{rl}\underset{R\to \mathrm{\infty }}{lim}\left|{\int }_{{\mathrm{\Gamma }}_R}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z\right|& \le \underset{R\to \mathrm{\infty }}{lim}\frac{\pi R^{3/2}}{(R-1)(\alpha R^2-1)}\\ & =0,\end{array}$$

which implies

$$\underset{R\to \mathrm{\infty }}{lim}{\int }_{{\mathrm{\Gamma }}_R}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z=0.$$

The integral along the straight path $L_2$ below the real axis may be evaluated as

$$\begin{array}{rl}\underset{R\to \mathrm{\infty }}{lim}\underset{ϵ\to 0}{lim}{\int }_{L_2}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z& =\underset{R\to \mathrm{\infty }}{lim}\underset{ϵ\to 0}{lim}{\int }_R^{ϵ}\frac{e^{i\pi }{x}^{3/2}}{2(x+1)(\alpha x^2+1)}\mathrm{d}x\\ & =\underset{R\to \mathrm{\infty }}{lim}\underset{ϵ\to 0}{lim}{\int }_{ϵ}^R\frac{x^{3/2}}{2(x+1)(\alpha x^2+1)}\mathrm{d}x\\ & =I^{\prime }(\alpha ).\end{array}$$

Bounding the integral along ${\gamma }_{ϵ}$ follows the same procedure as done for ${\mathrm{\Gamma }}_R$. We find that

$$\begin{array}{rl}\underset{ϵ\to 0}{lim}\left|{\int }_{{\gamma }_{ϵ}}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z\right|& \le \underset{ϵ\to 0}{lim}\frac{\pi {ϵ}^{3/2}}{(ϵ-1)(\alpha {ϵ}^2-1)}\\ & =0,\end{array}$$

which implies

$$\underset{ϵ\to 0}{lim}{\int }_{{\gamma }_{ϵ}}\frac{z\cdot z^{1/2}}{2(z+1)(\alpha z^2+1)}\mathrm{d}z=0.$$

Combining these results shows

$$\begin{array}{}\text{(3)}& \underset{R\to \mathrm{\infty }}{lim}\underset{ϵ\to 0}{lim}\frac{1}{2}{\oint }_{C_{R,ϵ}}\frac{z\cdot z^{1/2}}{(z+1)(\alpha z^2+1)}\mathrm{d}z& =2I^{\prime }(\alpha ).\end{array}$$

Making an equality between $(2)$ and $(3)$ gives

$$\begin{array}{rl}{I}^{\prime }(\alpha )& =\frac{\pi }{2\sqrt{2}}\cdot \frac{{\alpha }^{-3/4}-{\alpha }^{-1/4}}{1+\alpha }+\frac{\pi }{2}\cdot \frac{1}{1+\alpha },\text{for }\alpha >0.\end{array}$$

Then, by the way we conveniently placed $\alpha$ in the definition of $I(\alpha )$, we have

$$\begin{array}{rl}I& =I(1)-I(0)\\ & ={\int }_0^1{I}^{\prime }(\alpha )\mathrm{d}\alpha \\ & =\frac{\pi }{2\sqrt{2}}{\int }_0^1\frac{{\alpha }^{-3/4}-{\alpha }^{-1/4}}{1+\alpha }\mathrm{d}\alpha +\frac{\pi }{2}{\int }_0^1\frac{1}{1+\alpha }\mathrm{d}\alpha \\ & =\pi \sqrt{2}{\int }_0^1\frac{{\alpha }^{-3/4}-{\alpha }^{-1/4}}{1+\alpha }\mathrm{d}\alpha +\frac{\pi }{2}\ln (2)\\ & =\pi \sqrt{2}{\int }_0^1\frac{1-u^2}{1+u^4}\mathrm{d}u+\frac{\pi }{2}\ln (2),\alpha =u^4.\end{array}$$

We can then compute the antiderivative of this last integral as

$$\begin{array}{rl}\int \frac{1-x^2}{1+x^4}\mathrm{d}x& =\frac{1}{2\sqrt{2}}\int \frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1}\mathrm{d}x-\frac{1}{2\sqrt{2}}\int \frac{-2x+\sqrt{2}}{-x^2+\sqrt{2}x-1}\mathrm{d}x,\\ & \text{then let }{u}_1=x^2+\sqrt{2}x+1\text{ in the first integral and }\\ & \text{let }{u}_2=-x^2+\sqrt{2}x-1\text{ in the second integral to get}\\ & =\frac{\ln (|u_1|)-\ln (|u_2|)}{2\sqrt{2}}+C\\ & =\frac{\ln (|x^2+\sqrt{2}x+1|)-\ln (|-x^2+\sqrt{2}x-1|)}{2\sqrt{2}}+C.\end{array}$$

So

$$\begin{array}{rl}I& =\pi \sqrt{2}\left[\frac{\ln (|x^2+\sqrt{2}x+1|)-\ln (|-x^2+\sqrt{2}x-1|)}{2\sqrt{2}}\right]{|}_0^1+\frac{\pi }{2}\ln (2)\\ & =\pi \sqrt{2}\cdot \frac{\ln (3+2\sqrt{2})}{2\sqrt{2}}+\frac{\pi }{2}\ln (2)\\ & =\pi \ln (\sqrt{2}\cdot \sqrt{3+2\sqrt{2}})\\ & =\pi \ln \left(\sqrt{6+4\sqrt{2}}\right).\end{array}$$

Denesting the radical gives

$${\int }_1^{\mathrm{\infty }}\frac{\ln (x^4-2x^2+2)}{x\sqrt{x^2-1}}\mathrm{d}x=\pi \ln (2+\sqrt{2})$$

as desired.