Check if an integral is convergent and calculate its value

Question

I have improper integral $$\int_0^\pi \frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx$$

I have to check if it is convergent. If yes, then i have to evaluate it.

I think it is convergent. I can make substitution $t=\sin(x)$. But what next?

It might be interesting to make a substitution $t=1-2\sin x$... — abiessu, Sep 11 '15 at 14:42
Your substitution is also OK, but the one suggested by @abiessu may be simpler. You ask what next, what is the current integral you got? — mickep, Sep 11 '15 at 14:45
The issue is the behaviour near $\pi/6$ (and $5\pi/6$, but that's similar). To investigate, make the substitution suggested by abiessu. — André Nicolas, Sep 11 '15 at 14:52
The answer strongly depends on how you define the function $f(x)=x^{1/3}$ for negative $x$. — Jack D'Aurizio, Sep 11 '15 at 14:57

kleineg · Accepted Answer · 2015-09-11T15:07:00.670

1

First, since it is an improper integral you need to split it into the sum of three parts

$$\lim_{a\rightarrow\frac{5\pi}6}\lim_{b\rightarrow\frac{\pi}6}\int_a^\pi \frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx + \int_b^a\frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx + \int_0^b\frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx$$

If you make the substitution $t = 1-2\sin(x)$ you will get $dt = -2\cos(x) dx$.

The boundaries become $t(\pi) = 1-2\sin(\pi) = 1$, $t(\frac{5\pi}6) = 1-2\sin(\frac{5\pi}6) = 0$, $t(\frac{\pi}6) = 1-2\sin(\frac{\pi}6) = 0$, and $t(0) = 1-2\sin(0) = 1$

Use this to simplify your equation to $$\int_0^1\frac{-1}{2t^{1/3}} dt + \int_0^0\frac{-1}{2t^{1/3}} dt + \int_1^0\frac{-1}{2t^{1/3}} dt$$

We will use $$\int_0^0\frac{-1}{2t^{1/3}} dt = 0$$ and $\int_1^0\frac{-1}{2t^{1/3}} dt = -\int_0^1\frac{-1}{2t^{1/3}} dt$

So the sum totals to $0$

edited Sep 11 '15 at 15:07

answered Sep 11 '15 at 14:52

kleineg

1,795

Can I use substiution first and then split integral? – adm34 Sep 11 '15 at 15:12
I believe you can if the integral converges. That is the way I answered at first before deleting my answer and coming from this direction. – kleineg Sep 11 '15 at 15:17
if the upper limit of the integral was pi/2 instead of pi then integral would be non zero right? – adm34 Sep 11 '15 at 15:18
With the upper limit of $\frac\pi2$ you would simplify to $$\int_0^1\frac{-1}{2t^{1/3}} dt = -\frac34$$ – kleineg Sep 11 '15 at 15:24
Wolfram says it's $3/2$... – adm34 Sep 11 '15 at 15:38
I plugged it into Wolfram like this – kleineg Sep 11 '15 at 15:43
But the question is if we can do substitution like that to improepr integral? – adm34 Sep 11 '15 at 15:46
the factor of $-\frac12$ comes from the substitution. in essence you have $dx = \frac{-1}{2\cos(x)} dt $ – kleineg Sep 11 '15 at 15:46
That is the question. The short answer is yes, the long one is only if the integral converges. I split the integral into pieces, then used a substitution which left no singularities in the integrals. This led me to believe that the integral converges to $0$. If I had just done the substitution without checking convergence first, I would not be confident that the answer was correct. – kleineg Sep 11 '15 at 15:52
Thanks a lot for clarifying! – adm34 Sep 11 '15 at 15:58
You are welcome. – kleineg Sep 11 '15 at 17:00

Ángel Mario Gallegos · Answer 2 · 2015-09-11T15:28:51.047

Formally, we can state \begin{align} \int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{\pi/2}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx\\ &=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{y=\pi/2}^{y=0}\frac{\cos (\pi-y)}{[1-2\sin (\pi-y)]^{1/3}}d(\pi-y)\\ \end{align} Where $y=\pi-x$, then \begin{align} \int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx-\int_{\pi/2}^{0}\frac{-\cos y}{(1-2\sin y)^{1/3}}dy\\ &=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{\pi/2}^0\frac{\cos y}{(1-2\sin y)^{1/3}}dy\\ &=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx-\int_0^{\pi/2}\frac{\cos y}{(1-2\sin y)^{1/3}}dy\\ &=0 \end{align} This is true if $$\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx$$ converges.

By using $t=\sin x,\;\;dt=\cos x dx$ we have $$\int\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\int\frac{dt}{(1-2t)^{1/3}}=-\frac{3}{4}(1-2t)^{2/3}+C=-\frac{3}{4}(1-2\sin x)^{2/3}+C$$ Then $$\int_0^{\pi/6}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{b\to\frac{\pi}{6}^-}\int_0^{b}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{b\to\frac{\pi}{6}^-}-\frac{3}{4}(1-2\sin b)^{2/3}+\frac{3}{4}=\frac{3}{4},$$ $$\int_{\pi/6}^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{a\to\frac{\pi}{6}^+}\int_a^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\frac{3}{4}+\lim_{a\to\frac{\pi}{6}^+}\frac{3}{4}(1-2\sin a)^{2/3}=\frac{3}{4}$$ Then, the integral converges, also $$\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\frac{3}{2}$$ Therefore $$\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=0$$

Check if an integral is convergent and calculate its value

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