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!
Best Answer
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.