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Let's define the differential one form $\omega$ as: $$\omega =-x_1^2 x_2 \;dx_1+x_1 x_2 \log(1+e^{2x_2})\;dx_2$$ and the curve $\Gamma:=\{x\in\mathbb{R}^2\mid x_1^2+x_2^2=1\}$

Compute the integral of $\omega$ along the curve $\Gamma$ (in the trigonometric direction). What I did so far:

We can use Green-Riemann's theorem: $\int_{\partial K} \omega = \int_K d\omega$ $$d\omega = x_1^2+x_2\log(1+e^{2x_2})\;dx_1\wedge dx_2$$ So we can develop as follows: $$\underset{\Gamma}{\int}\omega = \underset{D_1}{\int}x_1^2+x_2\log(1+e^{2x_2})\;dx_1dx_2$$ I then tried integrating $$\int_{-1}^1\underset{-\sqrt{1-x_2^2}}{\overset{\sqrt{1-x_2^2}}{\int}}x_1^2+x_2\log(1+e^{2x_2})dx_1dx_2$$$$=\int_{-1}^1\left(2x_2\log(1+e^{2x_2})+\frac{2}{3}(1-x_2^2)\right)\cdot\sqrt{1-x_2^2} \;dx_2$$ But I am stuck here... Wolfram gives me $\frac{\pi}{2}$ as an answer but I have no idea how to get to that. I also tried not using G-R. theorem and integrating directly but the integral was much worst looking full of trig functions. Is there maybe another way to solve this problem ? Or just Help integrating the final expression ?

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    $\begingroup$ Here's a hint: how can you break $\log(1+e^{2x_2})$ into an even function plus an odd function? $\endgroup$ – Jason Aug 28 '20 at 1:51
  • $\begingroup$ @Jason Thanks a lot, I just tried it and it's working :) $\endgroup$ – Leo Aug 28 '20 at 12:18
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I'll post the result as an answer so maybe it can be usefull to someone in the future. So after the hint given I finally found how to integrate this using the following steps. Using Green-Riemann we can write $$\underset{\Gamma}{\int} \omega = \underset{D_1}{\int} d\omega=\underset{D_1}{\int}x_1^2+x_2\log(1+e^{2x_2})dx_1dx_2$$ \begin{gather}=\underset{D_1}{\int}x_1^2\;dx_1dx_2+\int_{-1}^1\underset{-\sqrt{1-x_2^2}}{\overset{\sqrt{1-x_2^2}}{\int}}x_2\log(1+e^{2x_2})dx_1dx_2\\ =\frac{\pi}{4}+\int_{-1}^1 2x_2\log(1+e^{2x_2})\sqrt{1-x_2^2}\;dx_2\end{gather} We can now notice that: $\log(1+e^{2x})=\log\big((e^{-x}+e^{x})e^x\big)=\log(e^{-x}+e^x)+x$

Substituing: $$=\frac{\pi}{4}+\int_{-1}^1 2x_2^2\sqrt{1-x_2^2}\;dx_2+\int_{-1}^1 \underbrace{2x_2\sqrt{1-x_2^2}\log(e^{-x_2}+e^{x_2})}_{=:f(x_2)}\;dx_2$$ Now since $f(-x)=-f(x)\implies\displaystyle \int_{-1}^1 f(x)\;dx = 0$ and so we are left with a much easier problem: $$=\frac{\pi}{4}+\int_{-1}^1 2x_2^2\sqrt{1-x_2^2}\;dx_2 = \frac{\pi}{4}+\frac{\pi}{4}$$ $$\implies \underset{\Gamma}{\int} \omega = \frac{\pi}{2}$$

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