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This integral is from integral

Find $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$$

I have get $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx=\int_0^{\infty}\tan\left(\frac1{\sqrt{x+1}}\right)\frac{\ln(1+\sqrt x)}{x}dx$$ Let$$\dfrac{1}{\sqrt{x+1}}=t$$ that $$I=\int_{0}^{1}\dfrac{\tan{t}\ln{\left(1+\sqrt{\frac{1}{t^2}-1}\right)}}{t-t^3}dt$$ This integral is have closed form ?

math110
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2 Answers2

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From your form you can write $$I=\int_{0}^{\infty}\tan\left(\frac{x}{\sqrt{x^{3}+x^{2}}}\right)\cdot\frac{\ln\left(1+\sqrt{x}\right)}{x}dx=\int_{-1}^{1}\frac{\tan\left(x\right)\ln\left(1+\sqrt{\frac{1}{x^{2}}-1}\right)}{x\sqrt{1-x^{2}}}\cdot\frac{1}{\sqrt{1-x^{2}}}dx$$

Now, using Chebyshev node you can write

$$\int_{-1}^{1}\frac{g(x)}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\pi}{n}\sum_{j=1}^{n}g(x_j)\qquad\text{where }x_j=\cos\left(\frac{2j-1}{2n}\pi\right)$$

In this case

$$g\left(x\right)=\frac{\tan\left(x\right)}{x\sqrt{1-x^{2}}}\ln\left(1+\sqrt{\frac{1}{x^{2}}-1}\right)$$

So you have

$$I\approx\frac{2\pi}{n}\sum_{j=1}^{n}\frac{\tan\left(\cos\left(\frac{2j-1}{2n}\pi\right)\right)}{\sin\left(\frac{2j-1}{n}\pi\right)}\cdot\ln\left(1+\left|\tan\left(\frac{2j-1}{2n}\pi\right)\right|\right)$$

Important: since $g(x)$ is not defined in $0$, use an even $n$. The larger $n$ is, the better you will approximate the integral

Math Attack
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$$\int_{0}^{\infty}\tan\left(\frac{1}{\sqrt{x+1}}\right)\frac{\log\left(1+\sqrt{x}\right)}{x}\,dx$$ $\frac{x}{\sqrt{x^3+x^2}}=\frac{1}{\sqrt{x+1}}, \quad x>0$ $$z=\sqrt{x},\quad x=z^2,\quad dx=2z\,dz,\quad z\in \mathbb{C}$$ Our contour will be a semicircle $\Gamma_R$ paremeterized by $z=Re^{i \theta}, 0\leqslant \theta \leqslant \pi$

$$\frac{1}{\sqrt{z^2+1}}=\frac{\pi}{2}+n\pi, \quad \sqrt{z^2+1}=\frac{2}{\pi\left(1+2n\right)}, \quad z^2+1=\frac{4}{\pi^2\left(4n^2+4n+1\right)},\quad z=\pm\frac{\sqrt{4-\pi^2\left(4n^2+4n+1\right)}}{\pi\left(1+2n\right)},\quad z_n=\Im\frac{\sqrt{\pi^2\left(1+2n\right)^2-4}}{\pi\left|1+2n\right|}\,\text{ for } n\geqslant0$$ $w\left(z\right)=e^{-\frac{1}{2}\log\left(z^2+1\right)}$ $$\lim_{R\to\infty}\oint_{\Gamma_R}\tan\left(w\left(z\right)\right)\frac{\log\left(1+z\right)}{z}\,dz= 2\pi\Im\,\text{res}\left(f;z_n\right)$$ $$\text{res}\left(f;z_n\right)= \sum_{n\geqslant 0}\frac{\log\left(z_n\right)}{z_n}$$

$$\int_{0}^{\infty}\tan\left(\frac{1}{\sqrt{x+1}}\right)\frac{\log\left(1+\sqrt{x}\right)}{x}\,dx = \Re\left[\pi \Im \sum_{n\geqslant 0}\frac{\log\left(z_n\right)}{z_n}\right]$$

Daniel K
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    Please make your answer more readable by adding explanations to each step and clarify what you are doing for each step. – Yunxuan Zhang Feb 16 '25 at 21:40