Slippery integral

Though I'm aware that the author of the question is looking for an 'elementary' solution, I will post my method just for reference and also to illustrate the power of complex analysis in cases like this one.

Let the given integral be $I$ .

Also note that,

$\int_{0}^{2\pi}e^{cosx}sin(sinx)dx=0$

Thus, $I=\Re (\int_{0}^{2\pi}e^{cosx}(cos(sinx)+isin(sinx))dx)$ $I=\Re (\int_{0}^{2\pi}e^{cosx}e^{isinx}dx)$ $I=\Re (\int_{0}^{2\pi}e^{cosx+isinx}dx)$ $I=\Re (\int_{0}^{2\pi}e^{e^{ix}}dx)$

Let us first evaluate this integral and $then$ find it's real part.

So let $e^{ix}=z$ .

Then it is easy to see that z always lies on a unit circle centered at the origin and our integral is now a contour integral in the complex plane along this unit circle.

Now, $ie^{ix}dx=dz$ and thus,

$I=\oint_{\left | z \right |=1}\frac{e^z}{iz}dz=(2\pi)\frac{1}{2\pi i} \oint_{\left | z \right |=1}\frac{e^z}{z}dz$

Finally, from Cauchy's integral formula ,

$I=2\pi \frac{\mathrm{d} }{\mathrm{d} z}_{z=0}(e^z)=2\pi$

This completes the proof. Note that the imaginary part of our integral turns out to be zero as expected.

Edit: Well, I seem to have found an 'elementary' method as well.

We start with what we already proved: $I=\Re (\int_{0}^{2\pi}e^{e^{ix}}dx)$

Now, $e^{z}=1+z+\frac{z^2}{2!}...$ is an identity that holds in the complex domain as well.

Using this fact: $I=\Re (\int_{0}^{2\pi}1+e^{ix}+\frac{(e^{ix})^2}{2!}...dx)$ $\Re(x+\frac{e^{ix}}{i}+\frac{e^{2ix}}{i*2*2!}+\frac{e^{3ix}}{i*3*3!}...)| ^{2\pi}_ {0}$ Thus, $I=2\pi$ . (As all the terms containing $e^{ix}$ evaluate to zero)

First use Eulers formula and convert given function to this :

$\displaystyle{f\left( x \right) =\cfrac { { e }^{ { e }^{ ix } }+{ e }^{ { -e }^{ ix } } }{ 2 } \\ A=\int _{ 0 }^{ 2\pi }{ \cfrac { { e }^{ { e }^{ ix } }+{ e }^{ { -e }^{ ix } } }{ 2 } dx } \\ f(x)=\cfrac { \left( 1+\cfrac { { e }^{ ix } }{ 1! } +\cfrac { { e }^{ 2ix } }{ 2! } ......\infty \right) +\left( 1+\cfrac { { e }^{ -ix } }{ 1! } +\cfrac { { e }^{ -2ix } }{ 2! } .....\infty \right) }{ 2 } \\ f(x)=1+\cfrac { \cos { x } }{ 1! } +\cfrac { \cos { 2x } }{ 2! } +\cfrac { \cos { 3x } }{ 3! } ......\infty \\ A=\int { dx } +\underbrace { \int { \cfrac { \cos { x } }{ 1! } dx+\cfrac { \cos { 2x } }{ 2! } dx+\cfrac { \cos { 3x } }{ 3! } dx......\infty } }_{ =0 } \\ A=\int _{ 0 }^{ 2\pi }{ f(x)dx } =2\pi }$

Q.E.D

---

---

$Proof$ :: Proof of this above integral , i.e $I=\int _{ 0 }^{ 2\pi }{ \cfrac { \cos { x } }{ 1! } dx+\cfrac { \cos { 2x } }{ 2! } dx+\cfrac { \cos { 3x } }{ 3! } dx......\infty } =0$

$\displaystyle{I=\int _{ 0 }^{ 2\pi }{ \sum _{ n=1 }^{ \infty }{ \cfrac { \cos { nx } }{ n! } } dx } =\sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ n! } \int _{ 0 }^{ 2\pi }{ \cos { nx } } dx } \\ I=\sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ n! } { \left[ \cfrac { \sin { nx } }{ n } \right] }_{ 0 }^{ 2\pi } } \\ \boxed { I=0 } \\ H.P}$

$AI^{2}TS ~X$ right. Don't know how to solve this