$\int_0^4\frac{\log x}{\sqrt{4x-x^2}} dx=0$

Question

Tengo problemas para demostrar que es igual a cero analíticamente. He probado a trazar y sé que para $0<x<1$ el integrando es negativo y positivo en caso contrario. He intentado la sustitución $u\to \sqrt{x}$ pero no puedo seguir adelante.

Answer 1

$$4x-x^2=2^2-(x-2)^2$$

Dejemos que $x-2=2\cos2t$

$$\int_0^4\dfrac{\ln(x)}{\sqrt{4x-x^2}}dx=-4\int_{\pi/2}^0\ln(2+2\cos2t)dt$$

$$=2\ln2\int_0^{\pi/2}dt+4\int_0^{\pi/2}\ln(\cos t)dt$$

Ver Evaluar $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$

Answer 2

Consideremos la integral $$I\left(s\right)=\int_{0}^{4}\frac{x^{s}}{\sqrt{4x-x^{2}}}dx,\, s>-1/2. $$ Observamos que $$ \int_{0}^{4}\frac{x^{s}}{\sqrt{4x-x^{2}}}dx\overset{x=4y}{=}4^{s+1}\int_{0}^{1}\frac{y^{s}}{\sqrt{16y-16y^{2}}}dy $$ $$=4^{s}\int_{0}^{1}y^{s-1/2}\left(1-y\right)^{-1/2}dy=4^{s}\frac{\Gamma\left(s+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(s+1\right)} $$ así que $$\frac{d}{ds}I\left(s\right)_{s=0}=\int_{0}^{4}\frac{\log\left(x\right)}{\sqrt{4x-x^{2}}}dx=\frac{d}{ds}\left(4^{s}\frac{\Gamma\left(s+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(s+1\right)}\right)_{s=0} $$ y observe que $$\begin{align} \frac{d}{ds}\left(4^{s}\frac{\Gamma\left(s+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(s+1\right)}\right)_{s=0}= & \left(\frac{4^{s}\Gamma\left(\frac{1}{2}\right)\Gamma\left(s+\frac{1}{2}\right)\left(\psi\left(s+\frac{1}{2}\right)-\psi\left(s+1\right)\log\left(4\right)\right)}{\Gamma\left(s+1\right)}\right)_{s=0} \\ = &0. \end{align} $$

Answer 3

Es posible demostrar que $$ \int_{0}^{4}\frac{\ln x}{\sqrt{4x-x^2}}\,dx=2\int_{0}^{\pi/2}\ln(2\sin u)\,du $$ Sí, es cierto, $$ \int_{0}^{4}\frac{\ln x}{\sqrt{4x-x^2}}\,dx= \int_{0}^{4}\frac{\ln x}{2\sqrt{1-(\frac{x-2}{2})^2}}\,dx \overset{t=\frac{x-2}{2}}{=}\int_{-1}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\ =\int_{-1}^{0}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt+\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\ \overset{u=-t}{=}\int_{1}^{0}\frac{\ln(2-2u)}{\sqrt{1-u^2}}\cdot-du+\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\ \int_{0}^{1}\frac{\ln(2-2t)}{\sqrt{1-t^2}}\,dt +\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\ $$ Ahora, $$ \int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\overset{t=\cos u}{=} \int_{\pi/2}^{0}\frac{\ln(2\cos u+2)}{\sin u}\cdot-\sin u du =\int_{0}^{\pi/2}\ln(2\cos u+2)\,du\\ =\int_{0}^{\pi/2}\ln(4\cos^2\frac{u}{2})\,du =2\int_{0}^{\pi/2}\ln(2\cos\frac{u}{2})\,du $$ de manera similar $$ \int_{0}^{1}\frac{\ln(2-2t)}{\sqrt{1-t^2}}\,dt =2\int_{0}^{\pi/2}\ln(2\sin\frac{u}{2})\,du $$ así que $$ \int_{0}^{4}\frac{\ln x}{\sqrt{4x-x^2}}\,dx =2\int_{0}^{\pi/2}\ln(2\sin u)\,du $$