calculation of $\int_{0}^{1}\tan^{-1}(1-x+x^2)dx$

Question

Compute the definite integral $$ \int_{0}^{1}\tan^{-1}(1-x+x^2)\,dx $$

Failed Attempt:

Let $1-x+x^2=t$. Then

$$ \begin{align} (2x-1)\,dx &= dt\\ dx &= \frac{1}{(2x-1)}dt \end{align} $$

Changing the limits of integration, we get

$$\int_{1}^{1}\tan^{-1}(t)\cdot \frac{1}{(2x-1)}dt = \int_{1}^{1}\tan^{-1}(t)\cdot f(t)dt = 0 $$

where $f(t)=\frac{1}{(2x-1)}$.

Is it true that $\int_{a}^{a}f(x)dx = 0$? If not, then where have I made a mistake in my attempted solution?

$\arctan(1-x+x^2)$ is positive on $[0,1]$, so certainly the integral is not zero. Also: for given $t$, equation $1-x+x^2=t$ has two solutions $x$. So do the two halves $[0,1/2]$ and $[1/2,1]$ separately, since $1-x+x^2$ is injective on each of those. — GEdgar, Dec 24 '13 at 03:20
Seems tangent has some really odd integrals. Every one of these questions contain tan/arctan. — cygorx, Dec 24 '13 at 06:07

score 20 · Accepted Answer · edited Apr 13 '17 at 12:21

20

HINT:

As $\displaystyle 1-x+x^2=\frac{(2x-1)^2+3}4>0$ for real $x,$

using this, $\displaystyle \tan^{-1}(1-x+x^2)=\cot^{-1}\left(\frac1{1-x+x^2}\right)$

Now, we know $\displaystyle\cot^{-1}\left(\frac1{1-x+x^2}\right)=\frac\pi2-\tan^{-1}\left(\frac1{1-x+x^2}\right)$

Again, $\displaystyle\tan^{-1}\left(\frac1{1-x+x^2}\right)=\tan^{-1}\left(\frac{x+1-x}{1-x(1-x)}\right)=\tan^{-1}x+\tan^{-1}(1-x)$

From this, the last identity holds true if $\displaystyle x(1-x)\le1$ which is true as $\displaystyle x(1-x)=\frac{1-(2x-1)^2}4\le \frac14$ for real $x$

Finally, as $\displaystyle \int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$

$\displaystyle \int_0^1\tan^{-1}(1-x)dx=\int_0^1\tan^{-1}\{1-(\underbrace{1+0-x})\}dx=\int_0^1\tan^{-1}xdx$

edited Apr 13 '17 at 12:21

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answered Dec 24 '13 at 03:20

lab bhattacharjee

274,582

wow, like a magic trick – Jonathan Dec 24 '13 at 03:34
1

@SalechAlhasov, thanks, please find the edited version – lab bhattacharjee Dec 24 '13 at 13:50
@Jonathan, thanks, I've added something more to the answer – lab bhattacharjee Dec 25 '13 at 12:27
@juantheron, Some related links : http://www.math.cmu.edu/~ploh/docs/math/3-telescope-solns.pdf, http://web.mit.edu/yisun/www/notes/trigtricks.pdf, http://math.stackexchange.com/questions/450352/infinite-series-telescoping and http://goo.gl/zb6pI0 – lab bhattacharjee Dec 25 '13 at 12:29
Thanks lab bhattacharjee.... – juantheron Dec 25 '13 at 14:22
@juantheron, my pleasure. Hope I could make the idea clear – lab bhattacharjee Dec 25 '13 at 14:30

score 8 · Answer 2 · answered Dec 24 '13 at 03:47

While lab bhattacharjee's hint is indeed excellent, it's also perhaps too high-brow for many students to notice unless pointed out to them, so I'd like to offer an alternative hint based on standard methods instead of "tricks".

Hint: Integrating by parts,

$$\int\tan^{-1}{\left(1-x+x^2\right)}\,dx=x\tan^{-1}{\left(1-x+x^2\right)}-\int x\cdot\frac{-1+2x}{\left(1-x+x^2\right)^2+1}dx\\ =x\tan^{-1}{\left(1-x+x^2\right)}-\int\frac{x\left(2x-1\right)}{\left(1-x+x^2\right)^2+1}dx.$$

Then consider using partial fractions to evaluate this new integral.

Luckily, that quartic factors into $(x^2+1)(x^2-2x+2)$. – Jonathan Dec 24 '13 at 03:55 — Jonathan, Dec 24 '13 at 03:55

calculation of $\int_{0}^{1}\tan^{-1}(1-x+x^2)dx$

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