How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$

Question

Let $L = \displaystyle \int_0^1 \frac{dx}{1+{x^8}}$ . Then

1. $L < 1$

2. $L > 1$

3. $L < \frac{\pi}{4}$

4. $L > \frac{\pi}{4}$

I got some idea from this video link. But got stuck while evaluating the second integral.

Please help!!

Thanks in advance!

Answer 1

$\forall x \in [0,1], \frac{1}{1+x^8}<1$, so $L<1$.

$\forall x \in [0,1], \frac{1}{1+x^8}>\frac{1}{1+x^2}$, so $L>\int_0^1 \frac{dx}{1+{x^2}}=\frac{\pi}{4}$

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How to evaluate this integral :

$$\begin{align} \int \frac{1}{1+x^8}dx &= \int (1+x^8)^{-1}dx \\&= \int_0^x(1+t^8)^{-1}dt +a \\&= \int_0^{x^8} (1+t)^{-1}d(t^{1/8})+b \\ &=\frac{1}{8}\int_0^{x^8}(1+t)^{-1}t^{1/8 -1}dt+c \\ &= \frac{1}{8}\int_0^{1} (1+x^8t)^{-1} (x^8t)^{1/8-1} d(x^8t)+d \tag{1} \\ &= \frac{x}{8} \int_0^1(1+x^8t)^{-1}x^{1-8}t^{1/8-1}x^8dt+e \tag{2} \\ &= x\,{}_2F_1 \left( 1,\frac{1}{8},\frac{9}{8},-x^8 \right) +f\end{align}$$

So $L={}_2F_1 \left( 1,\frac{1}{8},\frac{9}{8},-1 \right)$.

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$(1)$ I substitute $t$ with $x^8t$
$(2)$ Here $x$ is just a constant

Answer 2

There is a "closed-form" expression for this integral. Using partial fractions, $$ \dfrac{1}{1+x^8} = - \dfrac{1}{8} \sum_{\omega} \dfrac{\omega}{x - \omega}$$ where the sum is over the roots of the polynomial $x^8 + 1$. Then integrate: $$ \int_0^1 \dfrac{dx}{1+x^8} = -\dfrac{1}{8} \sum_\omega \omega (\log(1-\omega) - \log(-\omega))$$ where we use the principal branch of the logarithm (note that the line segments from $-\omega$ to $1-\omega$ never cross the negative real axis, which is the branch cut of this function, so this works). These roots can be expressed explicitly as $\left(\pm \sqrt {2+\sqrt {2}}\pm i\sqrt {2-\sqrt {2}}\right)/2$ and $\left(\pm \sqrt {2-\sqrt {2}}\pm i\sqrt {2+\sqrt {2}}\right)/2$. The final result is a rather messy, but explicit, expression in terms of $16$ complex logarithms. It can then be simplified to

$$\frac{1}{16} \left( \ln \left( \left( 4-2\,\sqrt {2} \right) \sqrt {2- \sqrt {2}}-4\,\sqrt {2}+7 \right) +\pi \right) \sqrt {2-\sqrt {2}}+\frac{1}{16} \left( \ln \left( \left( 4+2\,\sqrt {2} \right) \sqrt {2+\sqrt {2}}+4\,\sqrt {2}+7 \right) +\pi \right) \sqrt {2+\sqrt {2}}$$

The numerical value turns out to be approximately $.9246517057$.

Of course, this is not the best way to answer the original question.

Answer 3

You can get to a closed form by partial fractions without too much trouble, it just takes adherence to a protocol that keeps things relatively simple. First, expand into linear complex fractions $$\frac1{x^8+1}=\frac1{\prod_{k=0}^7(x-\omega_k)}=\sum_{k=0}^7\frac{A_k}{x-\omega_k}$$ Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$ and $\theta_k=\frac{\pi(2k+1)}8$. Then $$\lim_{x\rightarrow\omega_k}\frac{x-\omega_k}{x^8+1}=\lim_{x\rightarrow\omega_k}\frac1{8x^7}=\frac1{8\omega_k^7}=-\frac{\omega_k}8=\lim_{x\rightarrow\omega_k}\sum_{j=0}^7A_j\frac{x-\omega_k}{x-\omega_j}=\sum_{j=0}^7A_j\delta_{jk}=A_k$$ Just as @Robert Israel said. But then we diverge for a while. Here I would use the symmetry $$\theta_{7-k}=\frac{\pi(15-2k)}8=2\pi-\theta_k$$ So $\omega_{7-k}=\omega_k^*$. Then $$\begin{align}\frac1{x^8+1}&=-\frac18\sum_{k=0}^7\frac{\omega_k}{x-\omega_k}=-\frac18\sum_{k=0}^3\left(\frac{\omega_k}{x-\omega_k}+\frac{\omega_k^*}{x-\omega_k^*}\right)\\ &=-\frac18\sum_{k=0}^3\frac{2x\cos\theta_k-2}{x^2-2x\cos\theta_k+1}=-\frac14\sum_{k=0}^3\frac{(x-\cos\theta_k)\cos\theta_k-\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\end{align}$$ And now we can integrate more smoothly $$\begin{align}\int\frac{dx}{x^8+1}&=-\frac14\sum_{k=0}^3\int\frac{(x-\cos\theta_k)\cos\theta_k-\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\\ &=-\frac14\sum_{k=0}^3\left[\frac12\cos\theta_k\ln\left(x^2-2x\cos\theta_k+1\right)-\sin\theta_k\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$ Now we can use another symmetry $$\theta_{3-k}=\frac{\pi(7-2k)}8=\pi-\theta_k$$ So $\cos\theta_{3-k}=-\cos\theta_k$ and $\sin\theta_{3-k}=\sin\theta_k$. We can add the arctangents because $$\begin{align}\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)+\tan^{-1}\left(\frac{x+\cos\theta_k}{\sin\theta_k}\right)&=\tan^{-1}\left(\frac{\frac{x-\cos\theta_k}{\sin\theta_k}+\frac{x+\cos\theta_k}{\sin\theta_k}}{1-\frac{x^2-\cos^2\theta_k}{\sin^2\theta_k}}\right)\\ &=\tan^{-1}\left(\frac{2x\sin\theta_k}{1-x^2}\right)\end{align}$$ And subtracting logarithms leads to ratios, so $$\int_0^1\frac{dx}{x^8+1}=-\frac14\sum_{k=0}^1\left[\frac12\cos\theta_k\ln\left(\frac{x^2-2x\cos\theta_k+1}{x^2-2x\cos\theta_k+1}\right)-\sin\theta_k\tan^{-1}\left(\frac{2x\sin\theta_k}{1-x^2}\right)\right]_0^1$$ We can now use $$\cos\theta_0=\sin\theta_1=\frac{\sqrt{2+\sqrt2}}2$$ $$\cos\theta_1=\sin\theta_0=\frac{\sqrt{2-\sqrt2}}2$$ $$\lim_{x\rightarrow\infty}\tan^{-1}x=\frac{\pi}2$$ And $\tan0=\ln1=0$ to get to $$\begin{align}\int_0^1\frac{dx}{x^8+1}&=-\frac{\sqrt{2+\sqrt2}}{16}\ln\left(\frac{2-\sqrt{2+\sqrt2}}{2+\sqrt{2+\sqrt2}}\right)+\frac{\sqrt{2-\sqrt2}\cdot\pi}{16}\\ &-\frac{\sqrt{2-\sqrt2}}{16}\ln\left(\frac{2-\sqrt{2-\sqrt2}}{2+\sqrt{2-\sqrt2}}\right)+\frac{\sqrt{2+\sqrt2}\cdot\pi}{16}\end{align}$$ After checking that it matches the numerical result of $0.924651705775538$ we are satisfied with the result, same as @Robert Israel, but we have shown this time how the symmetries of the problem keep the expansions from getting out of hand. As has already been pointed out, this is between the easy bounds of $\frac{\pi}4$ and $1$.

EDIT: I just had to try and get the closed form to fit on a single line. $$\int_0^1\frac{dx}{x^8+1}=\frac{\sqrt{2+\sqrt2}}{16}\left(\sqrt8\sinh^{-1}\sqrt[4]2+\pi\right)+\frac{\sqrt{2-\sqrt2}}{16}\left(\sqrt8\cosh^{-1}\sqrt[4]2+\pi\right)$$ Or even $$\int_0^1\frac{dx}{x^8+1}=\frac{\sqrt{2+\sqrt2}}{8\sqrt2}\left(2\sinh^{-1}\sqrt[4]2+\pi\right)+\frac{\sqrt{2-\sqrt2}}{4\sqrt2}\cosh^{-1}\sqrt[4]2$$

Answer 4

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\begin{align} &\color{#f00}{\int_{0}^{1}{\dd x \over 1 + x^{8}}}\ \stackrel{x^{8}\ \to\ x}{=}\ {1 \over 8}\int_{0}^{1}{x^{-7/8} \over 1 + x}\,\dd x = {1 \over 8}\bracks{% \int_{0}^{1}{\dd x \over 1 + x} - \int_{0}^{1}{1 - x^{-7/8} \over 1 + x}\,\dd x} \\[3mm] = &\ {1 \over 8}\,\ln\pars{2} - {1 \over 8}\bracks{2\int_{0}^{1}{1 - x^{-7/8} \over 1 - x^{2}}\,\dd x - \int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x} \\[3mm] = &\ {1 \over 8}\,\ln\pars{2} - {1 \over 8}\bracks{\int_{0}^{1}{x^{-1/2} - x^{-15/16} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x} \\[3mm] = &\ {1 \over 8}\,\ln\pars{2} - {1 \over 8}\bracks{\int_{0}^{1}{1 - x^{-15/16} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x} \\[3mm] = &\ {1 \over 8}\,\bracks{\ln\pars{2} - \Psi\pars{{1 \over 16}} + \Psi\pars{\half} + \Psi\pars{{1 \over 8}} + \gamma} \\[3mm] = &\ \color{#f00}{{1 \over 8}\bracks{\Psi\pars{{1 \over 8}} - \Psi\pars{{1 \over 16}} - \ln\pars{2}}} \approx 0.9247 \end{align}

where $\Psi$ is the Digamma function and $\gamma$ is the Euler-Mascheroni constant and with the use of the well known identities $$ \left\lbrace\begin{array}{rcl} \ds{\Psi\pars{z} + \gamma} & \ds{=} & \ds{\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\qquad\Re\pars{z} > 0} \\[1mm] \ds{\Psi\pars{\half}} & \ds{=} & \ds{-2\ln\pars{2} - \gamma} \end{array}\right. $$

A further reduction can be found with the identity ( see Gradshteyn & Ryzhik table ): \begin{align} \Psi\pars{{p \over q}} & = - \gamma - \ln\pars{2q} - {\pi \over 2}\,\cot\pars{\pi\,{p \over q}} + 2\sum_{k = 1}^{\left\lfloor\pars{q + 1}/2\right\rfloor - 1} \cos\pars{2\pi k\,{p \over q}}\ln\pars{\sin\pars{{k\pi \over q}}} \\[3mm] &\ q = 2,3,\ldots\,;\qquad p = 1,2,\ldots,\pars{q - 1} \end{align}

Answer 5

The best and fastest way is to express the integrand as the sum of an infinite geometric series where $r=-x^8$, which converges in domain of the integral. Since the question asks to approximate $L$, use the first couple terms of the series: \begin{align*} L& \approx \int_0^1 1-x^8 \; \mathrm{d}x \\ &= x-\frac{x^9}{9} \bigg \rvert_0^1 \\ &=\frac{8}{9} \approx 0.9 \\ \end{align*} The answer choices $L<1$ and $L>\frac{\pi}{4}$ are both correct.