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Evaluate the following integral :

$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$

I was tried use change variable ,

If I use $x=y^2$ integral becomes :

$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$

From here I have one idea the derivative under sing integral but I got I difficult integration :

$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$

I already to see you hints or solution!

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  • $\begingroup$ Mathematica gives: $$\pi \left(\log \left(2 \left(4+\sqrt{2}+2 \sqrt{2 \left(2+\sqrt{2}\right)}\right)\right)-\log \left(4 \left(\sin \left(\frac{\pi }{8}\right)-1\right) \left(\cos \left(\frac{\pi }{8}\right)-1\right)\right)\right)$$. I wouldn't spend time on this working "by hand"... but that's your call. $\endgroup$ Mar 22, 2020 at 23:01
  • $\begingroup$ @DavidG.Stork thank you sir , very difficult closed form $\endgroup$ Mar 22, 2020 at 23:04

3 Answers 3

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Let $$I(a)=\int_0^{\infty}\frac{\ln(a^8+x^4)}{\sqrt x(1+x)}dx=2\int_0^{\infty}\frac{\ln(a^8+x^8)}{1+x^2}$$ Then $$I(0)=2\int_0^{\infty}\frac{\ln(x^8)}{1+x^2}dx=2\int_{\infty}^0\frac{\ln(y^{-8})}{1+\frac1{y^2}}\left(-\frac{dy}{y^2}\right)=-2\int_0^{\infty}\frac{\ln(y^8)}{1+y^2}dy=-I(0)=0$$ And $$\begin{align}I^{\prime}(a)&=2\cdot8a^7\int_0^{\infty}\frac{dx}{(a^8+x^8)(1+x^2)}dx=8a^7\int_{-\infty}^{\infty}\frac{dx}{(a^8+x^8)(1+x^2)}\\ &=2\pi i\cdot8a^7\left(\frac1{(a^8+1)(2i)}+\sum_{n=0}^3\frac1{8a^7e^{\pi i(2n+1)\cdot7/8}(1+a^2e^{\pi i(2n+1)/4})}\right)\\ &=\frac{8\pi a^7}{a^8+1}+2\pi i\sum_{n=0}^3\frac{-e^{-\pi i(2n+1)/8}}{a^2+e^{-\pi i(2n+1)/4}}\\ &=\frac{8\pi a^7}{a^8+1}-\frac{2\pi i}{(-2i)}\sum_{n=0}^3\left(\frac1{a+ie^{-\pi i(2n+1)/8}}-\frac1{a-ie^{-\pi i(2n+1)/8}}\right)\end{align}$$ So $$\begin{align}I(1)&=I(0)+\int_0^1I^{\prime}(a)da\\ &=\pi\int_0^1\frac{8a^7}{a^8+1}da+\pi\sum_{n=0}^3\int_0^1\left(\frac1{a+ie^{-\pi i(2n+1)/8}}-\frac1{a-ie^{-\pi i(2n+1)/8}}\right)da\\ &=\pi\ln2+\left.\pi\sum_{n=0}^3\left(\ln\left(a+ie^{-\pi i(2n+1)/8}\right)-\ln\left(a-ie^{-\pi i(2n+1)/8}\right)\right)\right|_0^1\\ &=\pi\ln2+\pi\ln\left(\frac{\cos^2\frac{\pi}{16}\cos^2\frac{3\pi}{16}}{\sin^2\frac{\pi}{16}\sin^2\frac{3\pi}{16}}\right)\\ &=\pi\ln2+\pi\ln\left(\frac{\left(1+\cos\frac{\pi}8\right)\left(1+\cos\frac{3\pi}8\right)}{\left(1-\cos\frac{\pi}8\right)\left(1-\cos\frac{3\pi}8\right)}\right)\\ &=\pi\ln2+\pi\ln\left(\frac{\left(2+\sqrt{2+\sqrt2}\right)\left(2+\sqrt{2-\sqrt2}\right)}{\left(2-\sqrt{2+\sqrt2}\right)\left(2-\sqrt{2-\sqrt2}\right)}\right)\\ &=2\pi\ln\left(\left(2+\sqrt{2+\sqrt2}\right)\left(2+\sqrt{2-\sqrt2}\right)\right)\\ &=4\pi\ln\left(\sqrt2+\sqrt{2+\sqrt2}\right)\end{align}$$ WolframAlpha seems to agree with this result at least numerically.

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  • $\begingroup$ Very nice solution sir .......+1,thanks $\endgroup$ Mar 23, 2020 at 16:05
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Using the principal branch of the logarithm, we have

$$ \begin{align} I &= \int_{0}^{\infty} \frac{\ln(1+t^{4})}{\sqrt{t}(1+t)} \, \mathrm dt = 2\int_{0}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx = \int_{-\infty}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{7} \int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx + \int_{-\infty}^{\infty} \frac{\ln \left(1 \, {\color{red}{+}} \, xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \right)\\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx+ \int_{\infty}^{-\infty} \frac{\ln \left(1 - ue^{i \pi(2n+1)/8}\right)}{1+u^2} \, (- \mathrm du) \right) \\&= 2 \sum_{n=0}^{3} \int_{-\infty}^{\infty} \frac{\ln \left(1 - xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ & \stackrel{(1)}= 2\sum_{n=0}^{3} 2 \pi i \operatorname{Res} \left[ \frac{\ln \left(1-ze^{i \pi(2n+1)/8}\right)}{1+z^2}, i \right] \\ &= 2\sum_{n=0}^{3} 2 \pi i \, \frac{\ln \left(1-ie^{i \pi(2n+1)/8}\right)}{2i} \\ &= 2\pi \sum_{n=0}^{3} \ln \left(1-ie^{i \pi(2n+1)/8}\right) \\ &= 2 \pi \ln \left[ \left(1-ie^{i \pi/8} \right) \left(1-ie^{3 \pi i/8} \right) \left(1+ie^{-3 \pi i/8} \right) \left(1+ie^{- i \pi/8} \right) \right]\\ &= 2 \pi \ln \left[\left(2+ 2 \sin \left(\frac{\pi}{8} \right) \right) \left(2+ 2 \sin \left(\frac{3\pi}{8} \right) \right)\right] \\ &= 2 \pi \ln \left(4 +2 \sqrt{2-\sqrt{2}} +2 \sqrt{2+\sqrt{2}} + \sqrt{2} \right) \\ &= 2 \pi \ln \left(4 + 2 \sqrt{4 +2\sqrt{2}} + \sqrt{2}\right) \\ &= 2 \pi \ln \left(\left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right)^{2}\right) \\ &= 4 \pi \ln \left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right) \end{align}$$


$(1)$ The branch cut for $\ln \left(1-ze^{i \pi(2n+1)/8}\right) $ is in the lower half of the complex plane.

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Note $$1+x^8=(1+e^{i \frac\pi4}x^2)(1+e^{-i \frac\pi4}x^2)(1+e^{i \frac{3\pi}4}x^2)(1+e^{-i \frac{3\pi}4}x^2)$$

$$J(a)=\int_0^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= \pi\ln(1+a^{\frac12}) $$ Then \begin{align} I&=\int\limits_0^{\infty}\frac{\ln (1+x^{4})}{\sqrt{x}(1+x)}dx\overset{x\to x^2} =2\int\limits_0^{\infty}\frac{\ln (1+x^{8})}{1+x^{2}}dx\\ &= 2\left[J(e^{i \frac\pi4})+ J(e^{-i \frac\pi4})+J(e^{i \frac{3\pi}4}) +J(e^{-i \frac{3\pi}4}) \right]\\ &=2\pi \ln\left[(1+e^{i \frac\pi8})(1+e^{-i \frac\pi8})(1+e^{i \frac{3\pi}8})(1+e^{-i \frac{3\pi}8})\right]\\ &=2\pi \ln\left[4(1+\cos\frac\pi8)(1+\cos\frac{3\pi}8) \right]\\ &=4\pi\ln (\sqrt2+\sqrt{2+\sqrt2}) \end{align}

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