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Here's my attempt at an integral I found on this site. $$\int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=2\pi$$ I'm not asking for a proof, I just want to know where I messed up

Recall that, for all $x$, $$e^x=\sum_{n\geq0}\frac{x^n}{n!}$$ And $$\cos x=\sum_{n\geq0}(-1)^n\frac{x^{2n}}{(2n)!}$$ Hence we have that $$ \begin{align} \int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=&\int_0^{2\pi}\bigg(\sum_{n\geq0}\frac{\cos^n2x}{n!}\bigg)\bigg(\sum_{m\geq0}(-1)^m\frac{\sin^{2m}2x}{(2m)!}\bigg)\mathrm{d}x\\ =&\sum_{n,m\geq0}\frac{(-1)^m}{n!(2m)!}\int_0^{2\pi}\cos(2x)^n\sin(2x)^{2m}\mathrm{d}x\\ =&\frac12\sum_{n,m\geq0}\frac{(-1)^m}{n!(2m)!}\int_0^{4\pi}\cos(t)^n\sin(t)^{2m}\mathrm{d}t\\ \end{align} $$ The final integral is related to the incomplete beta function, defined as $$B(x;a,b)=\int_0^x u^{a-1}(1-u)^{b-1}\mathrm{d}u$$ If we define $$I(x;a,b)=\int_0^x\sin(t)^a\cos(t)^b\mathrm{d}t$$ We can make the substitution $\sin^2t=u$, which gives $$ \begin{align} I(x;a,b)=&\frac12\int_0^{\sin^2x}u^{a/2}(1-u)^{b/2}u^{-1/2}(1-u)^{-1/2}\mathrm{d}u\\ =&\frac12\int_0^{\sin^2x}u^{\frac{a-1}2}(1-u)^{\frac{b-1}2}\mathrm{d}u\\ =&\frac12\int_0^{\sin^2x}u^{\frac{a+1}2-1}(1-u)^{\frac{b+1}2-1}\mathrm{d}u\\ =&\frac12B\bigg(\sin^2x;\frac{a+1}2,\frac{b+1}2\bigg)\\ \end{align} $$ Hence we have a form of our final integral: $$ \begin{align} I(4\pi;2m,n)=&\frac12B\bigg(\sin^24\pi;\frac{2m+1}2,\frac{n+1}2\bigg)\\ =&\frac12B\bigg(0;\frac{2m+1}2,\frac{n+1}2\bigg)\\ =&\frac12\int_0^0t^{\frac{2m-1}2}(1-t)^{\frac{n-1}2}\mathrm{d}t\\ =&\,0 \end{align} $$ Which implies that $$\int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=0$$ Which is totally wrong. But as far as I can tell, I haven't broken any rules. Where's my error, and how do I fix it? Thanks.

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You cannot substitute $u=\sin^2t$. As $t$ ranges from $0$ to $2\pi$, this is not a one-to-one relationship.

It's like if you subbed $u=x^2$ in $$\int_{-1}^1x^2\,dx$$ You would get an integral from $u=1$ to $u=1$, which would be $0$ even though the integral is clearly nonzero.

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  • $\begingroup$ What do you recommend I do instead? $\endgroup$ – clathratus Dec 1 '18 at 4:41
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    $\begingroup$ I don't know. The question explicitly asks for "where I messed up". It's not asking for the right way to calculate this integral. $\endgroup$ – alex.jordan Dec 1 '18 at 4:42
  • $\begingroup$ You're right. Thank you. $\endgroup$ – clathratus Dec 1 '18 at 4:43
  • $\begingroup$ @clathratus I have not read you solution but in case of an even function $f(x)$, $\int_{-a}^{+a}f(x)dx=2\int_0^af(x)dx$ $\endgroup$ – tatan Dec 1 '18 at 4:59
  • $\begingroup$ What does $\cos \sin 2 x$ mean? $\endgroup$ – David G. Stork Dec 1 '18 at 7:00
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To solve the integral, you may consider $$\int_{C} \frac{e^z}{z}dz$$ where $C$ is a unit circle, and see its real part.

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If you want a full solution of the integral using complex analysis, going along the lines of what Seewoo Lee recommend, you can solve the integral as follows:

First consider the integral $$\int_{C} \frac{e^z}{z}dz$$

where $C$ is the unit circle oriented counter-clockwise in the complex plane. Using the Cauchy Integral formula (from complex analysis) you can find that $$\int_{C} \frac{e^z}{z}dz = 2\pi i \phantom{------}(1)$$

Now we will directly integrate the above integral by parametrising $C$. Let $z(t)=e^{2it}$ with $0 \le t \le \pi$ be the parametrisation of $C$. Then the integral works out as:

\begin{align} \int_{C} \frac{e^z}{z}dz &= \int_0^{\pi} \frac{e^{e^{2it}}}{e^{2it}} 2ie^{2it} dt \\ &= 2i \int_0^{\pi} {e^{\cos(2t)+i\sin(2t)}}dt \\ &= 2i \int_0^{\pi} {e^{\cos(2t)}e^{i\sin(2t)}}dt \\ &= 2i \int_0^{\pi} {e^{\cos(2t)}(\cos(\sin(2t))+i\sin(\sin(2t))} dt \\ &= 2i \int_0^{\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt - 2 \int_0^{\pi} {e^{\cos(2t)}}(sin(sin(2t)) dt \phantom{-------} (2) \\ \end{align}

Equating imaginary parts of (1) and (2) we see that:

$$\int_0^{\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt = \pi$$

If we parametrise the curve initially with $\pi \le t \le 2\pi$ we would have ended up with:

$$\int_{\pi}^{2\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt = \pi$$

Thus,

\begin{align} \int_{0}^{2\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt &= \int_{0}^{\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt + \int_{\pi}^{2\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt \\ &= \pi + \pi = 2\pi \end{align}

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you could try this: $$I=\int_0^{2\pi}e^{\cos(2x)}\cos\left[\sin(2x)\right]dx$$ $$=\Re\left(\int_0^{2\pi}e^{\cos(2x)}\cos\left[\sin(2x)\right]dx+i\int_0^{2\pi}e^{\cos(2x)}\sin\left[\sin(2x)\right]dx\right)$$ $$=\Re\left(\int_0^{2\pi}e^{\cos(2x)}e^{i\sin(2x)}dx\right)$$ $$=\Re\left(\int_0^{2\pi}e^{\cos(2x)+i\sin(2x)}dx\right)$$ $$=\Re\left(\int_0^{2\pi}e^{e^{2ix}}dx\right)$$ and since: $$e^y=\sum_{n=0}^\infty\frac{x^n}{n!}$$ we can say that: $$e^{e^{2ix}}=\sum_{n=0}^\infty\frac{e^{2nix}}{n!}$$ and so: $$I=\Re\int_0^{2\pi}\sum_{n=0}^\infty\frac{e^{2nix}}{n!}dx$$ $$=\Re\sum_{n=0}^\infty\left[\frac{e^{2nix}}{2ni.n!}\right]_0^{2\pi}$$ $$=\Re\sum_{n=0}^\infty\frac{e^{4\pi ni}-1}{2ni.n!}$$ but note that for all integers n, $$e^{4\pi ni}=1$$ so this summation may be hard to calculate (or wrong). This is probably due to the fact that the integral is between $0$ and $2\pi$, so the integral may need to be split up into several parts before it can be evaulated.

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  • $\begingroup$ Neat approach for sure! Thanks for sharing (+1). $\endgroup$ – clathratus Dec 2 '18 at 19:03
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If by $\cos \sin (2 x)$ you really mean $\cos (2 x ) \sin (2 x)$, then the full function looks like

enter image description here

and the full integral is $0$.

If instead the integral is:

$$\int\limits_{x=0}^{2 \pi} e^{\cos (2 x)} \cos \left( \sin ( 2 x)\right)\ dx$$

the graph is:

enter image description here

and Mathematica gives the answer as $2 \pi$.

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    $\begingroup$ I believe that the function is $\cos(\sin(2x))$. $\endgroup$ – robjohn Dec 1 '18 at 7:24
  • $\begingroup$ Really? Then the OP is not very careful with mathematical notation. $\endgroup$ – David G. Stork Dec 1 '18 at 7:41
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    $\begingroup$ I prefer to write $\cos(x)$, but many write $\cos x$ even though that often leads to confusion. However, inserting an extra $2x$ into the formula to get $\cos(2x)\sin(2x)$ seems questionable. $\endgroup$ – robjohn Dec 1 '18 at 8:18
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    $\begingroup$ I did mean $\cos(\sin(2x))$. Sorry for the confusion. I appreciate the thoroughness of your answer though (+1) $\endgroup$ – clathratus Dec 1 '18 at 20:13

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