Re: Integral
Posted: Thu Jul 14, 2016 7:09 am
by Papapetros Vaggelis
Hi Tolis.
The function \(\displaystyle{f:\left[0,\dfrac{1}{\sqrt{2}}\right]\longrightarrow \mathbb{R}}\) is well defined and continuous at
\(\displaystyle{\left[0,\dfrac{1}{\sqrt{2}}\right]}\), so : \(\displaystyle{\int_{0}^{\frac{1}{\sqrt{2}}}f(x)\,\mathrm{d}x<\infty}\).
Consider the indefinite integral \(\displaystyle{I=\int \dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x}\) at \(\displaystyle{\left[0,\dfrac{1}{\sqrt{2}}\right]}\).
\(\displaystyle{\int \dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x=\int \dfrac{\sqrt{1+x^4}}{\left(1-x^2\right)\,\left(1+x^2\right)}\,\mathrm{d}x}\).
By substituting \(\displaystyle{t=\arctan\,x\,,t\in\left[0,\arctan\,\dfrac{1}{\sqrt{2}}\right]\subseteq \left[0,\dfrac{\pi}{4}\right)}\), we get :
\(\displaystyle{\mathrm{d}t=\dfrac{1}{1+x^2}\,\mathrm{d}x}\) and \(\displaystyle{x=\tan\,t}\), so :
\(\displaystyle{\begin{aligned}I&=\int \dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x\\&=\int \dfrac{\sqrt{1+\tan^4\,t}}{1-\tan^2\,t}\,\mathrm{d}t\\&=\int \dfrac{\displaystyle{\sqrt{\dfrac{\sin^4\,t+\cos^4\,t}{\cos^4\,t}}}}{\displaystyle{\dfrac{\cos^2\,t-\sin^2\,t}{\cos^2\,t}}}\,\mathrm{d}t\\&=\int \dfrac{\sqrt{\left(\sin^2\,t+\cos^2\,t\right)^2-2\,\sin^2\,t\,\cos^2\,t}}{\cos\,(2\,t)}\,\mathrm{d}t\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\sqrt{2-\sin^2\,(2\,t)}}{1-\sin^2\,(2\,t)}\,2\,\cos\,(2\,t)\,\mathrm{d}t\end{aligned}}\)
Now, setting \(\displaystyle{u=\sin\,(2\,t)\,,u\in\left[0,1\right)}\), we have that : \(\displaystyle{\mathrm{d}u=2\,\cos\,(2\,t)\,\mathrm{d}t}\) and thus :
\(\displaystyle{I=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\sqrt{2-u^2}}{1-u^2}\,\mathrm{d}u}\).
By applying the substitution \(\displaystyle{u=\sqrt{2}\,\sin\,y\,,y\in\left[0,\dfrac{\pi}{4}\right)}\), we get : \(\displaystyle{\mathrm{d}u=\sqrt{2}\,\cos\,y\,\mathrm{d}y}\) and then :
\(\displaystyle{\begin{aligned} I&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\sqrt{2-2\,\sin^2\,y}}{1-2\,\sin^2\,y}\,\sqrt{2}\,\cos\,y\,\mathrm{d}y\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{2\,\cos^2\,y}{\cos\,(2\,y)}\,\mathrm{d}y\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \dfrac{\cos\,(2\,y)+1}{\cos\,(2\,y)}\,\mathrm{d}y\\&=\dfrac{1}{2\,\sqrt{2}}\,\int \left(1+\dfrac{1}{\cos\,(2\,y)}\right)\,\mathrm{d}y\end{aligned}}\)
where \(\displaystyle{\int 1\,\mathrm{d}y=y+c\,,c\in\mathbb{R}}\) and
\(\displaystyle{\begin{aligned} \int \dfrac{1}{\cos\,(2\,y)}\,\mathrm{d}y&=\int \dfrac{\cos\,(2\,y)}{\cos^2\,(2\,y)}\,\mathrm{d}y\\&=\dfrac{1}{2}\,\int \dfrac{1}{1-\sin^2\,(2\,y)}\,\mathrm{d}(\sin\,(2\,y))\\&=\dfrac{1}{4}\,\int \dfrac{\left(1-\sin\,(2\,y)\right)+\left(1+\sin\,(2\,y)\right)}{\left(1-\sin\,(2\,y)\right)\,\left(1+\sin\,(2\,y)\right)}\,\mathrm{d}(\sin\,(2\,y))\\&=\dfrac{1}{4}\,\int \left[\dfrac{1}{1+\sin\,(2\,y)}+\dfrac{1}{1-\sin\,(2\,y)}\right]\,\mathrm{d}(\sin\,(2\,y))\\&=\dfrac{1}{4}\,\ln\,\left|\dfrac{1+\sin\,(2\,y)}{1-\sin\,(2\,y)}\right|+c'\,,c'\in\mathbb{R}\end{aligned}}\)
so :
\(\displaystyle{I=y+\dfrac{1}{4}\,\ln\,\left|\dfrac{1+\sin\,(2\,y)}{1-\sin\,(2\,y)}\right|+c\,,c\in\mathbb{R}}\).
Now, we have that :
\(\displaystyle{u=\sqrt{2}\,\sin\,y \implies u^2=2\,\sin^2\,y=1-\cos\,(2\,y)\implies u^2-1=-\cos\,(2\,y)\implies \left(u^2-1\right)^2=\cos^2\,(2\,y)=1-\sin^2\,(2\,y)}\)
or : \(\displaystyle{\sin^2\,(2\,y)=1-\left(u^2-1\right)^2\iff \sin\,(2\,y)=\sqrt{1-\left(u^2-1\right)^2}=\sqrt{u^2\,\left(2-u^2\right)}=u\,\sqrt{2-u^2}}\).
Also, \(\displaystyle{y=\arcsin\,\dfrac{u}{\sqrt{2}}=\arcsin\,\dfrac{\sin\,(2\,t)}{\sqrt{2}}=\arcsin\,\dfrac{\sin\,(2\,\arctan\,x)}{\sqrt{2}}}\).
\(\displaystyle{u=\sin\,(2\,t)=\sin\,(2\,\arctan\,x)}\), so :
\(\displaystyle{u^2-1=\sin^2\,(2\,\arctan\,x)-1=-\cos^2\,(2\,\arctan\,x)}\) and then :
\(\displaystyle{\sin\,(2\,y)=\sqrt{1-\left(u^2-1\right)^2}=\sqrt{1-\cos^4\,(2\,\arctan\,x)}=\sin\,(2\,\arctan\,x)\,\sqrt{1+\cos^2\,(2\,\arctan\,x)}}\).
In conclusion : $$ I=\frac{1}{2\sqrt{2}}\left [ \arcsin\frac{\sin \left ( 2\arctan x \right ) }{\sqrt{2}} +\frac{1}{4}\ln \left | \frac{1+\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2 \left ( 2\arctan x \right )}}{1-\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2\left ( 2\arctan x \right )}} \right | \right ]+c, \; c \in \mathbb{R}$$ and those functions are continuous and differentiable at \(\displaystyle{\left[0,\dfrac{1}{\sqrt{2}}\right]}\). Then :
\(\displaystyle{\int_{0}^{\frac{1}{\sqrt{2}}}\,\dfrac{\sqrt{1+x^4}}{1-x^4}\,\mathrm{d}x=}\) $$=\left [ \frac{1}{2\sqrt{2}}\left [ \arcsin\frac{\sin \left ( 2\arctan x \right ) }{\sqrt{2}} +\frac{1}{4}\ln \left | \frac{1+\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2 \left ( 2\arctan x \right )}}{1-\sin \left ( 2\arctan x \right )\sqrt{1+\cos^2\left ( 2\arctan x \right )}} \right | \right ] \right ] _0^{1/\sqrt{2}}$$
where :
\(\displaystyle{\sin\,(2\,\arctan\,x)=2\,\sin\,(\arctan\,x)\,\cos\,(\arctan\,x)}\) and
\(\displaystyle{\cos\,(\arctan\,x)=\dfrac{1}{\sqrt{1+\tan^2\,(\arctan\,x)}}=\dfrac{1}{\sqrt{1+x^2}}}\) ,
and then, since \(\displaystyle{\sin\,(\arctan\,x)>0\,,0\leq x\leq \dfrac{1}{\sqrt{2}}} \), we have that :
\(\displaystyle{\sin\,(\arctan\,x)=\sqrt{1-\cos^2\,(\arctan\,x)}=\dfrac{x}{\sqrt{1+x^2}}}\).
\(\displaystyle{I=\dfrac{1}{2\,\sqrt{2}}\,\left(f_{c}\left(\dfrac{1}{\sqrt{2}}\right)-f_{c}(0)\right)}\) gives the desired result.