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Here is my attempt. The result is not right. Please help identify the issue(s).

$\displaystyle f(x)=\int\cfrac{1}{x^4+1}\>\mathrm{d}x$, let $x=\tan t$, we have $ \mathrm{d}x = \sec^2 t\>\mathrm{d}t,\> t=\tan^{-1} x\in\left(-\cfrac{\pi}{2},\cfrac{\pi}{2}\right)$ \begin{align} \displaystyle f(\tan t)&= \int\cfrac{\sec^2 t\> \mathrm{d}t}{1+\tan^4 t}=\int\cfrac{\cos^2 t\> \mathrm{d}t}{\cos^4 t+\sin^4 t}=\int\cfrac{\cfrac{1+\cos 2t}{2}\> \mathrm{d}t}{(\cos^2 t+\sin^2 t)^2-2\sin^2 t\cos^2 t} \notag\\ &=\int\cfrac{1+\cos 2t}{2-\sin^2 2t} \>\mathrm{d}t =\int\cfrac{\mathrm{d}t}{2-\sin^2 2t} + \cfrac 12\int\cfrac{\mathrm{d}\sin 2t}{2-\sin^2 2t} \notag\\ &=\int\cfrac{\sec^2 2t \>\mathrm{d}t}{2\sec^2 2t-\tan^2 2t} + \cfrac {\sqrt{2}}8\int\cfrac{1}{\sqrt{2}-\sin 2t} + \cfrac{1}{\sqrt{2}+\sin 2t}\>\mathrm{d}\sin 2t \notag\\ &=\cfrac 12\int\cfrac{\mathrm{d}\tan 2t}{2+\tan^2 2t} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t} \notag\\ &=\cfrac {\sqrt{2}}4 \tan^{-1} \cfrac{\tan 2t}{\sqrt{2}} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t} \notag \end{align}

As $\tan 2t=\cfrac{2\tan t}{1-\tan^2 t}=\cfrac{2x}{1-x^2}, \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}=\cfrac{\sqrt{2}\sec^2 t+\tan t}{\sqrt{2}\sec^2 t-\tan t}=\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x},$

$f(x)=\cfrac {\sqrt{2}}4 \tan^{-1} \cfrac{\sqrt{2}x}{1+x^2} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+c$

If the above holds, $\displaystyle \int_0^{\infty} \cfrac{\mathrm{d} x}{1+x^4}$ would be $0$, which is impossible(Should be $\cfrac {\sqrt{2}\pi}{4}$).

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  • $\begingroup$ If I recall correctly, this is one of those integrals best approaches with an integration by parts (with $dv=1$). From there a change of variables should be straightforward. $\endgroup$
    – user123641
    Aug 6 '18 at 2:51
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    $\begingroup$ $x^4+1=x^4+2x^2+1-2x^2=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}+1)$ and you can use partial fractions $\endgroup$
    – saulspatz
    Aug 6 '18 at 2:55
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    $\begingroup$ Another approach is to use the fact that $$1/(x⁴+1)=\frac{1}{(x²+i)(x²-i)}=\frac{1}{(x+\sqrt{-i})(x-\sqrt{-i})(x+\sqrt{i})(x-\sqrt{i})}$$ and use partial fractions if you have the energy to work with ugly numbers $\endgroup$
    – ℋolo
    Aug 6 '18 at 2:58
  • $\begingroup$ @Holo but we should try to avoid complex no.s, right ?? I think it doesn't help in any way. $\endgroup$ Aug 6 '18 at 3:27
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Comparing to the method I used in the following, maybe the issue occurs when computing $$ \int \frac {\mathrm dt} {2-\sin^2(2t)}. $$ Then from now on $t$ cannot take the value $\pm \pi /4$ if we want to devide the numerator and the denominator by $\cos^2(2t)$. Now to compute the improper integral, we should take the limit $x \to 1^-$ and $x\to 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute $$ f(+\infty) - f(1^+) + f(1^-) - f(0), $$ which would give the right result $\sqrt 2 \pi/4$.

Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.

Appendix

I'm here to give another approach. We would introduce a conjugate pair. Assume $x \neq 0$. \begin{align*} \int \frac {\mathrm d x} {1+x^4} &= \frac 12 \int \frac {1-x^2}{1+x^4} \mathrm dx + \int \frac {1+x^2}{1+x^4}\mathrm dx \\ &= \frac 12 \int \frac {x^{-2} - 1}{x^2+x^{-2 }} \mathrm dx + \frac 12 \int \frac {x^{-2} + 1}{x^2+x^{-2 }} \mathrm dx\\ &= -\frac 12 \int \frac {\mathrm d (x + x^{-1})} {(x+x^{-1} )^2 -2} \mathrm dx+\frac 12 \int \frac {\mathrm d(x-x^{-1})} {(x - x^{-1})^2 +2}\\ &= -\frac {\sqrt 2}8 \int \left( \frac 1 {x +x^{-1}-\sqrt 2} - \frac 1{x+x^{-1}+ \sqrt 2}\right)\mathrm d(x+x^{-1}) \\ &\phantom{==}+\frac {\sqrt2}4 \int \frac {\mathrm d(x - x^{-1})/\sqrt 2} {((x-x^{-1})/\sqrt 2)^2 +1} \\ &= \frac {\sqrt 2}8 \log \left( \frac {x + x^{-1}+\sqrt 2} {x +x^{-1}-\sqrt 2}\right) + \frac {\sqrt 2}4 \mathrm {arctan}\left( \frac {x-x^{-1}} {\sqrt 2}\right) + C \\ &= \frac {\sqrt 2}8 \log \left(\frac {x^2 +\sqrt 2 x + 1} {x^2 - \sqrt2 x +1}\right) +\frac {\sqrt 2}4 \mathrm{arctan} \left(\frac {x^2 -1}{\sqrt 2 x}\right) + C. \end{align*}

If we use this as the result, then $$ f(+\infty) - f(0) = \frac {\sqrt 2} 4 \left( \frac \pi 2 + \frac \pi 2\right) = \frac {\sqrt 2}4 \pi. $$

Also note that when $x \neq 0$, $$ \arctan(x) + \mathrm{arccot} (x) = \mathrm {sgn} (x)\frac \pi 2 \implies \arctan (x) = \mathrm {sgn} (x)\frac \pi 2 + \arctan \left(-\frac 1x\right), $$ so the OP is correct.

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First of all, you made a typo in the final answer — the correct answer must be $$f(x)=\frac{\sqrt{2}}{4}\tan^{-1}\frac{\sqrt{2}x}{1\color{red}{-}x^2}+\frac{\sqrt{2}}{8}\ln\frac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C.$$

The next issue is the introduction of $\sec(2t)$ and $\tan(2t)$ when you switched to $$\int\frac{\sec^2 2t\,\mathrm{d}t}{2\sec^2 2t-\tan^2 2t}$$ (as part of an expression). Both $\sec(2t)$ and $\tan(2t)$ are undefined at some points within the domain $\displaystyle t\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, viz. at $\displaystyle t=\pm\frac{\pi}{4}$. Therefore, the antiderivative you find in terms of $t$ is in fact a piecewise-defined function: $$f(x(t))=\begin{cases} \cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\tan2t}{\sqrt{2}}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}+C_1, \text{ if } t\in\left(-\cfrac{\pi}{2},-\cfrac{\pi}{4}\right); \\ \cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\tan2t}{\sqrt{2}}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}+C_2, \text{ if } t\in\left(-\cfrac{\pi}{4},\cfrac{\pi}{4}\right); \\ \cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\tan2t}{\sqrt{2}}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}+C_3, \text{ if } t\in\left(\cfrac{\pi}{4},\cfrac{\pi}{2}\right). \end{cases}$$

Switching back to $x$ still creates a piecewise-defined function: $$f(x)=\begin{cases} \cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\sqrt{2}x}{1\color{red}{-}x^2}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C_1, \text{ if } x\in(-\infty,-1); \\ \cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\sqrt{2}x}{1\color{red}{-}x^2}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C_2, \text{ if } x\in(-1,1); \\ \cfrac{\sqrt{2}}{4}\tan^{-1}\cfrac{\sqrt{2}x}{1\color{red}{-}x^2}+\cfrac{\sqrt{2}}{8}\ln\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+C_3, \text{ if } x\in(1,+\infty). \end{cases}$$

At the points $x=\pm1$, these expressions are undefined, and so the corresponding integrals have to be treated as improper. In your case, the integral $\displaystyle \int_0^{+\infty}$ has to be split at the discontinuity at $x=1$: $$\int_0^{+\infty}\cdots\,\mathrm{d}x=\int_0^1\cdots\,\mathrm{d}x+\int_1^{+\infty}\cdots\,\mathrm{d}x,$$ and then, when evaluating the antiderivative that you found, you'll have to take the one-sided limits from the left and from the right at $x=1$, which are NOT equal to each other! And that's probably the source of your wrong answer.

More specifically: $$\lim_{x\to1^{-}}\frac{\sqrt{2}x}{1-x^2}=+\infty \implies \lim_{x\to1^{-}}\arctan\frac{\sqrt{2}x}{1-x^2}=\frac{\pi}{2},$$ while $$\lim_{x\to1^{+}}\frac{\sqrt{2}x}{1-x^2}=-\infty \implies \lim_{x\to1^{-}}\arctan\frac{\sqrt{2}x}{1-x^2}=-\frac{\pi}{2}.$$

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    $\begingroup$ This problem of branch cuts can be fixed by writing $\arctan(\sqrt{2}x +1) + \arctan(\sqrt{2}x - 1)$ instead of $\arctan[\sqrt{2}x/(1-x^2)]$, which gives the correct function. $\endgroup$ Aug 6 '18 at 4:32
  • $\begingroup$ @eyeballfrog: I agree. However, the OP's question wasn't asking for a different solution, but to help find out where the proposed solution went wrong. So I specifically limited my answer to an analysis of the proposed solution only. $\endgroup$
    – zipirovich
    Aug 6 '18 at 4:43
  • $\begingroup$ @zipirovich That's very considerate. I was having hard time to pick the "correct answer". Both of you pointed out my error. I chose yours over xbh's because you pointed out how the error is introduced. Thank you! $\endgroup$
    – Lance
    Aug 6 '18 at 13:47

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