I'm having trouble with this equation in my calculus course.
Show that: $$\int_{0}^{\inf}\frac{dx}{1+x^4}\le \frac{4}{3}$$
I've tried to use the comparison test (unsure if it's called that in English). The equation I tried to compare it to was $\int_{0}^{\inf}\frac{dx}{x^4}$ but that integral does not converge given the limits. I've also considered splitting the integral into smaller sections, but that has not worked either.
Thankful for any help!
On
It is possible to calculate this integral.
Let us define $$ I=\int_{0}^{+\infty} \frac{\mathrm{d} t}{1+t^4},\quad J=\int_{0}^{+\infty} \frac{t^2}{1+t^4}\,\mathrm{d} t $$
The function $t\mapsto \frac{1}{1+t^4}$ is continuous on $\mathbb{R}_{+}$ and $\frac{1}{1+t^4}=\mathrm{O}(\frac{1}{t^4})$ as $t$ approaches $+\infty$, we deduce that $\int_{0}^{+\infty} \frac{\mathrm{d} x}{1+x^4}$ exists by the criterion of domination and by comparison to a convergent Riemann integral.
We perform the change of variable $t=1/u$ in the integral $I$. The change of variable is valid since the mapping $u\mapsto 1/u$ is a strictly decreasing bijective $\mathscr{C}^1$ function between $\mathbb{R}_{+}^{*}$ and $\mathbb{R}_{+}^{*}$. We obtain $$ I = -\int_{+\infty}^{0} \frac{1/u^2}{1+1/u^4}\,\frac{1}{u^{2}}\,\mathrm{d} u = \int_{0}^{+\infty} \frac{u^{2}}{u^4+1}\,\mathrm{d} u = J. $$ The function $u\mapsto \mathrm{e}^{u}$ is a strictly increasing $\mathscr{C}^1$ bijection from $\mathbb{R}$ to $]0,{+\infty}[$, thus the change of variable $t=\mathrm{e}^{u}$ is valid. By making this change of variable in $I$, we obtain an integral of the same nature, and therefore convergent: $$ I = \int_{-\infty}^{+\infty} \frac{\mathrm{e}^{u}}{1+\mathrm{e}^{4u}}\,\mathrm{d} u \quad \text{and similarly}\quad J = \int_{-\infty}^{+\infty} \frac{\mathrm{e}^{3u}}{1+\mathrm{e}^{4u}}\,\mathrm{d} u. $$ Since $I=J$ we have $$ 2\,I = \int_{-\infty}^{+\infty} \frac{\mathrm{e}^{u}+\mathrm{e}^{3u}}{1+\mathrm{e}^{4u}}\,\mathrm{d} u = \int_{-\infty}^{+\infty} \frac{\mathrm{e}^{-u}+\mathrm{e}^{u}}{\mathrm{e}^{-2u}+\mathrm{e}^{2u}}\,\mathrm{d} u = \int_{-\infty}^{+\infty} \frac{\mathrm{ch}(u)}{\mathrm{ch}(2u)}\,\mathrm{d} u = \int_{-\infty}^{+\infty} \frac{\mathrm{ch}(u)}{1+2\mathrm{sh}(u)^2}\,\mathrm{d} u $$ then $$ 2\,I = \frac{1}{\sqrt{2}}\int_{-\infty}^{+\infty} \frac{\sqrt{2}\mathrm{ch}(u)}{1+(\sqrt{2}\mathrm{sh}(u))^2}\,\mathrm{d} u = \frac{1}{\sqrt{2}}\Bigl[\arctan\bigl(\sqrt{2}\mathrm{sh}(u)\bigr)\Bigr]_{-\infty}^{+\infty} = \frac{1}{\sqrt{2}}\Bigl(\frac{\pi}{2}+\frac{\pi}{2}\Bigr) = \frac{\pi}{\sqrt{2}} $$ and finaly $I=\frac{\pi}{2\sqrt{2}}=\frac{\pi\sqrt{2}}{4}$.
We have $$ \frac{\pi\sqrt{2}}{4}\leq \frac{4}{3} \quad \Longleftrightarrow\quad \frac{2\pi^2}{16}\leq \frac{16}{9} \quad \Longleftrightarrow\quad \pi^2\leq \frac{128}{9}. $$ We have $\pi\leq \frac{7}{2}$ therefore $\pi^2\leq \frac{49}{4}\leq \frac{128}{9}$. Finaly $$ \int_{0}^{+\infty} \frac{\mathrm{d} x}{1+x^4}\leq \frac{4}{3}. $$
$\int_0^{1}\frac 1 {1+x^{4}} dx \leq \int_0^{1}1dx =1$.
$\int_{1}^{\infty}\frac 1 {1+x^{4}} dx \leq \int_{1}^{\infty}\frac 1 {x^{4}} dx=\frac 1 3$.