Calculus: improper integration If $\displaystyle L = \int_0^1 \dfrac{dx}{(1+x^8)} $, then what is the upper bound and lower bound for $L$. I tried with numerical methods. But not got the answer
 A: If you decompose $$\dfrac{1}{(1+x^8)}$$ in partial fractions (which is not the most pleasnt thing to do), you should end (after a lot of simplifcations) with the awful thing
$$\displaystyle L = \int \dfrac{dx}{(1+x^8)}=\frac{1}{8} \left(\cos \left(\frac{\pi }{8}\right) \left(-\log \left(x^2-2 x \cos
   \left(\frac{\pi }{8}\right)+1\right)+\log \left(x^2+2 x \cos \left(\frac{\pi
   }{8}\right)+1\right)+2 \tan ^{-1}\left(x \sec \left(\frac{\pi
   }{8}\right)-\sqrt{2}+1\right)+2 \tan ^{-1}\left(x \sec \left(\frac{\pi
   }{8}\right)+\tan \left(\frac{\pi }{8}\right)\right)\right)+\sin \left(\frac{\pi
   }{8}\right) \left(-\log \left(x^2-2 x \sin \left(\frac{\pi
   }{8}\right)+1\right)+\log \left(x^2+2 x \sin \left(\frac{\pi
   }{8}\right)+1\right)-2 \tan ^{-1}\left(x \left(-\csc \left(\frac{\pi
   }{8}\right)\right)+\sqrt{2}+1\right)+2 \tan ^{-1}\left(x \csc \left(\frac{\pi
   }{8}\right)+\sqrt{2}+1\right)\right)\right)$$ and for the integral
$$\frac{1}{8} \left(\sin \left(\frac{\pi }{8}\right) \left(\pi +2 \tanh ^{-1}\left(\sin
   \left(\frac{\pi }{8}\right)\right)\right)+\cos \left(\frac{\pi }{8}\right)
   \left(\pi +2 \tanh ^{-1}\left(\cos \left(\frac{\pi
   }{8}\right)\right)\right)\right)$$
Added later to this answer
Beside this rigorous solution, you can represent your integral as an infinite series. Start with $$\frac{1}{1+y}=\sum_{i=0}^\infty (-1)^{(i)} y^i$$ and replace $y$ by $x^8$. So $$\int \dfrac{dx}{(1+x^8)}=\sum_{i=0}^\infty \frac{(-1)^{i} y^{8i+1}}{8i+1}$$ and $$\int_0^1 \dfrac{dx}{(1+x^8)}=\sum_{i=0}^\infty \frac{(-1)^{i}}{8i+1}$$ The partial sum for the last term is given by $$\sum_{i=0}^{n} \frac{(-1)^{i}}{8i+1}=\frac{1}{16} \left(2 (-1)^n \Phi \left(-1,1,n+\frac{9}{8}\right)+\pi  \csc
   \left(\frac{\pi }{8}\right)+4 \cos \left(\frac{\pi }{8}\right) \log \left(\cot
   \left(\frac{\pi }{16}\right)\right)+4 \sin \left(\frac{\pi }{8}\right) \log
   \left(\cot \left(\frac{3 \pi }{16}\right)\right)\right)$$ where appears the Lerch function. Going to limits, we end with $$\sum_{i=0}^\infty \frac{(-1)^{i}}{8i+1}=\frac{1}{16} \left(\pi  \csc \left(\frac{\pi }{8}\right)+4 \cos \left(\frac{\pi
   }{8}\right) \log \left(\cot \left(\frac{\pi }{16}\right)\right)+4 \sin
   \left(\frac{\pi }{8}\right) \log \left(\cot \left(\frac{3 \pi
   }{16}\right)\right)\right)$$ which is in fact strictly identical to the previously given result.
You can have a nice generalization of the above for the calculation of $$I_k=\int_0^1 \dfrac{dx}{(1+x^k)}=\sum_{i=0}^\infty \frac{(-1)^{i}}{k ~i+1}=\frac{\Phi \left(-1,1,\frac{1}{k}\right)}{k}$$ which leads do $$I_1=\log (2)$$ $$I_2=\frac{\pi }{4}$$ $$I_3=\frac{1}{9} \left(\sqrt{3} \pi +\log (8)\right)$$ $$I_4=\frac{\pi +2 \sinh ^{-1}(1)}{4 \sqrt{2}}$$ $$I_6=\frac{1}{6} \left(\pi +2 \sqrt{3} \coth ^{-1}\left(\sqrt{3}\right)\right)$$ $$I_8=\frac{1}{16} \left(\pi  \csc \left(\frac{\pi }{8}\right)+4 \cos \left(\frac{\pi
   }{8}\right) \log \left(\cot \left(\frac{\pi }{16}\right)\right)+4 \sin
   \left(\frac{\pi }{8}\right) \log \left(\cot \left(\frac{3 \pi
   }{16}\right)\right)\right)$$ $$I_{10}=\frac{1}{20} \left(\pi +\sqrt{5} \pi +\sqrt{2 \left(5+\sqrt{5}\right)} \log
   \left(\cot \left(\frac{\pi }{20}\right)\right)+\sqrt{10-2 \sqrt{5}} \log
   \left(\cot \left(\frac{3 \pi }{20}\right)\right)\right)$$ $$I_{12}=\frac{2 \left(1+\sqrt{3}\right) \pi +\log \left(577+408 \sqrt{2}\right)+\sqrt{3} \log
   \left(49+20 \sqrt{6}\right)}{24 \sqrt{2}}$$ $$$$
A: Following Lucian's hint, I am trying to give some details
$\forall x\in (0,1)$ we have $x^2>x^8\Rightarrow \frac{1}{1+x^2}<\frac{1}{1+x^8}\Rightarrow \displaystyle \int_0^1\frac{dx}{1+x^8}>\displaystyle \int_0^1\frac{dx}{1+x^2}=\tan^{-1}(1)-\tan^{-1}(0)=\frac{\pi}{4}$
And also $1+x^8>1\Rightarrow \displaystyle \int_0^1\frac{dx}{1+x^8}<\displaystyle \int_0^1dx=1$
A: Hint: $0<x^8<x^2$, for $x\in(0,1)$.
