Ytukyg

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

Recall that, for all $x$,
$$e^x=sum_{ngeq0}frac{x^n}{n!}$$
And $$cos x=sum_{ngeq0}(-1)^nfrac{x^{2n}}{(2n)!}$$
Hence we have that
$$
begin{align}
int_0^{2pi}e^{cos2x}cos(sin2x) mathrm{d}x=&int_0^{2pi}bigg(sum_{ngeq0}frac{cos^n2x}{n!}bigg)bigg(sum_{mgeq0}(-1)^mfrac{sin^{2m}2x}{(2m)!}bigg)mathrm{d}x\
=&sum_{n,mgeq0}frac{(-1)^m}{n!(2m)!}int_0^{2pi}cos(2x)^nsin(2x)^{2m}mathrm{d}x\
=&frac12sum_{n,mgeq0}frac{(-1)^m}{n!(2m)!}int_0^{4pi}cos(t)^nsin(t)^{2m}mathrm{d}t\
end{align}
$$
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
$$I(x;a,b)=int_0^xsin(t)^acos(t)^bmathrm{d}t$$
We can make the substitution $sin^2t=u$, which gives
$$
begin{align}
I(x;a,b)=&frac12int_0^{sin^2x}u^{a/2}(1-u)^{b/2}u^{-1/2}(1-u)^{-1/2}mathrm{d}u\
=&frac12int_0^{sin^2x}u^{frac{a-1}2}(1-u)^{frac{b-1}2}mathrm{d}u\
=&frac12int_0^{sin^2x}u^{frac{a+1}2-1}(1-u)^{frac{b+1}2-1}mathrm{d}u\
=&frac12Bbigg(sin^2x;frac{a+1}2,frac{b+1}2bigg)\
end{align}
$$
Hence we have a form of our final integral:
$$
begin{align}
I(4pi;2m,n)=&frac12Bbigg(sin^24pi;frac{2m+1}2,frac{n+1}2bigg)\
=&frac12Bbigg(0;frac{2m+1}2,frac{n+1}2bigg)\
=&frac12int_0^0t^{frac{2m-1}2}(1-t)^{frac{n-1}2}mathrm{d}t\
=&,0
end{align}
$$
Which implies that
$$int_0^{2pi}e^{cos2x}cos(sin2x) mathrm{d}x=0$$
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.

You cannot substitute $u=sin^2t$. As $t$ ranges from $0$ to $2pi$, 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.

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.

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 $$int_{C} frac{e^z}{z}dz = 2pi i phantom{------}(1)$$

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)+isin(2t)}}dt \
&= 2i int_0^{pi} {e^{cos(2t)}e^{isin(2t)}}dt \
&= 2i int_0^{pi} {e^{cos(2t)}(cos(sin(2t))+isin(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 $pi le t le 2pi$ we would have ended up with:

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

Thus,

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

you could try this:
$$I=int_0^{2pi}e^{cos(2x)}cosleft[sin(2x)right]dx$$
$$=Releft(int_0^{2pi}e^{cos(2x)}cosleft[sin(2x)right]dx+iint_0^{2pi}e^{cos(2x)}sinleft[sin(2x)right]dxright)$$
$$=Releft(int_0^{2pi}e^{cos(2x)}e^{isin(2x)}dxright)$$
$$=Releft(int_0^{2pi}e^{cos(2x)+isin(2x)}dxright)$$
$$=Releft(int_0^{2pi}e^{e^{2ix}}dxright)$$
and since:
$$e^y=sum_{n=0}^inftyfrac{x^n}{n!}$$
we can say that:
$$e^{e^{2ix}}=sum_{n=0}^inftyfrac{e^{2nix}}{n!}$$
and so:
$$I=Reint_0^{2pi}sum_{n=0}^inftyfrac{e^{2nix}}{n!}dx$$
$$=Resum_{n=0}^inftyleft[frac{e^{2nix}}{2ni.n!}right]_0^{2pi}$$
$$=Resum_{n=0}^inftyfrac{e^{4pi ni}-1}{2ni.n!}$$
but note that for all integers n, $$e^{4pi ni}=1$$
so this summation may be hard to calculate (or wrong). This is probably due to the fact that the integral is between $0$ and $2pi$, so the integral may need to be split up into several parts before it can be evaulated.

If by $cos sin (2 x)$ you really mean $cos (2 x ) sin (2 x)$, then the full function looks like

and the full integral is $0$.

If instead the integral is:

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

the graph is:

and Mathematica gives the answer as $2 pi$.