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Evaluation of inproper integral $\displaystyle \int^{\infty}_{-\infty}\frac{1}{x^4+64}dx$

Note that you need to use triangle inequality in your calculation.

What i try::

Let $$I=\int\frac{1/x^2}{x^2+64/x^2}dx=\frac{1}{8}\int\frac{8/x^2}{x^2+64/x^2}dx$$

$$I=\frac{1}{8}\int \frac{(8/x^2+1)+(8/x^2-1)}{x^2+64/x^2}dx$$

$$I=\frac{1}{8}\int\frac{8/x^2+1}{(x-8/x)^2-16}dx-\frac{1}{8}\int\frac{1-8/x^2}{(x-8/x)^2+16}dx$$

Put $\displaystyle x-8/x=u$ and $\displaystyle x+8/x=v$

So $$I=\frac{1}{8}\int\frac{1}{u^2-16}du-\frac{1}{8}\int\frac{1}{v^2+16}dv$$

Now abover convert into standard form.

I have a doubt on that problem.

I did not understand the meaning of use triangle inequality in your calculation.

Please explain me. Thanks.

jacky
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    If they are asking you to use the triangle inequality I would suggest they want you to bound the integral not evaluate it. Maybe I'm wrong, but I cannot see how the triangle inequality will help you evaluate that integral. –  Aug 12 '20 at 03:50
  • Means we have to find convergence of integration. Also plesse explain me How can i use triangle ineuality for bounds. Thanks – jacky Aug 12 '20 at 03:51
  • I did not understand How can i use Triangle inequality Here. Plaese Help me. – jacky Aug 12 '20 at 03:59
  • I don't see where the triangle inequality comes in. The answer given is probably a better approach. –  Aug 12 '20 at 04:02

3 Answers3

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$$\begin{eqnarray*} 8\int_{-\infty}^{+\infty}\frac{dx}{x^4+64}&\stackrel{x\to 2t}{=}&\int_{-\infty}^{+\infty}\frac{dt}{t^4+4}\stackrel{\text{Sophie Germain}}{=}\int_{-\infty}^{+\infty}\frac{dt}{(t^2+2t+2)(t^2-2t+2)}\\&\stackrel{\text{PFD}}{=}&\frac{1}{8}\int_{-\infty}^{+\infty}\frac{t+2}{(t+2)t+2}-\frac{t-2}{(t-2)t+2}\,dt\\&=&\frac{1}{8}\int_{-\infty}^{+\infty}\frac{(t+1)}{(t+1)^2+1}-\frac{(t-1)}{(t-1)^2+1}+\frac{1}{(t+1)^2+1}+\frac{1}{(t-1)^2+1}\,dt\\&=&\frac{1}{8}\int_{-\infty}^{+\infty}f(t+1)-f(t-1)\,dt+\frac{1}{4}\int_{-\infty}^{+\infty}\frac{dt}{t^2+1}\\&=&\frac{\pi}{4}+\int_{-\infty}^{+\infty}f(t+1)-f(t-1)\,dt\end{eqnarray*}$$ The triangle inequality can be employed in the last step, for showing that $\int_{-\infty}^{+\infty}f(t+1)-f(t-1)\,dt=0$ even if $f(t)=\frac{t}{t^2+1}$ does not belong to $L^1(\mathbb{R})$. Indeed $$ \int_{A}^{B}f(t+1)-f(t-1)\,dt = \int_{A+1}^{B+1}f(t)\,dt - \int_{A-1}^{B-1}f(t)\,dt $$ equals (assuming $A+1\leq B-1$) $$ \int_{B-1}^{B+1}f(t)\,dt - \int_{A-1}^{A+1}f(t)\,dt $$ which is bounded in absolute value by $ 4 \max |f(x)| $ over the set $|x|\geq \min(|B-1|,|A+1|)$.

Jack D'Aurizio
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  • What is "Sophie Germain" :-)? – Sebastiano Aug 12 '20 at 18:57
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    @Sebastiano: Marie-Sophie Germain was a great mathematician who lived in the era of Fermat and Euler. Couples $(p,2p+1)$ where both $p$ and $2p+1$ are primes take her name (the infinity of such couples is more or less equivalent to proving there are infinite twin primes); the identity $x^4+4=(x^2+2x+2)(x^2-2x+2)$ also take her name. – Jack D'Aurizio Aug 12 '20 at 19:01
  • I'm a scarce mathematician of an high school :-(. My sincerre compliments for your knowledges. – Sebastiano Aug 12 '20 at 19:02
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Not clear how you can use the triangle inequality. However, a different approach could use that $$ x^4+64 = (x^2 - 4x + 8) (x^2 + 4x + 8) $$ so you can apply partial fractions and get $$ \int_\mathbb{R} \frac{dx}{x^4+64} = \int_0^\infty \frac{2dx}{x^4+64} = \int_0^\infty \left(\frac{A+Bx}{x^2 - 4x + 8} + \frac{C+Dx}{x^2 + 4x + 8} \right)dx $$ where the denominators won't factor anymore but could be looked at as $(x\pm 2)^2 + 2^2$ and fit into standard form.

gt6989b
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Note

\begin{align} \int^{\infty}_{-\infty}\frac{dx }{x^4+64} &\overset{t=\frac x{2\sqrt2}} =\frac{\sqrt2}{16}\int_0^\infty \frac{dt}{t^4+1} \overset{t\to\frac 1t}=\frac{\sqrt2}{16}\cdot\frac12\int_0^\infty \frac{1+t^2 }{t^4+1}dt \\ & = \frac{\sqrt2}{32}\int_0^\infty \frac{d(t-\frac1t)}{(t-\frac1t)^2+2}= \frac{\sqrt2}{32}\cdot \frac\pi{\sqrt2} = \frac{\pi}{32} \end{align}

Quanto
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