MHB Definite Integral challenge #3

[Saitama]

Evaluate the following:
$$\int_0^{\pi} e^{\cos x} \cos(\sin x)\,\,dx$$

 

[alyafey22]

Evaluate the following:
$$\int_0^{\pi} e^{\cos x} \cos(\sin x)$$

Are you sure of the question because you are missing $dx$ ?

 

[Saitama]

Are you sure of the question because you are missing $dx$ ?

Haha, nice catch! Edited.

 

[polygamma]

Spoiler

$$\int_{0}^{\pi} e^{\cos x} \cos(\sin x) \ dx = \frac{1}{2} \text{Re} \int_{-\pi}^{\pi} e^{e^{iz}} \ dz = \frac{1}{2} \text{Re} \frac{1}{i} \int_{|z|=1} \frac{e^{z}}{z} \ dz $$

$$ = \frac{1}{2} \text{Re} \frac{1}{i} 2 \pi i \ \text{Res} \left[ \frac{e^{z}}{z},0 \right] = \text{Re} \ \pi (1) = \pi $$

 

[Saitama]

Spoiler

$$\int_{0}^{\pi} e^{\cos x} \cos(\sin x) \ dx = \frac{1}{2} \text{Re} \int_{-\pi}^{\pi} e^{e^{iz}} \ dz = \frac{1}{2} \text{Re} \frac{1}{i} \int_{|z|=1} \frac{e^{z}}{z} \ dz $$

$$ = \frac{1}{2} \text{Re} \frac{1}{i} 2 \pi i \ \text{Res} \left[ \frac{e^{z}}{z},0 \right] = \text{Re} \frac{1}{i} \pi i(1) = \pi $$

Thank you for your participation, your answer is correct. :)

 

[Saitama]

I forgot to mention in my previous post that there is still an alternative way which does not use the "res" thing used by Random Variable. The challenge is still open. :)

 

[I like Serena]

I forgot to mention in my previous post that there is still an alternative way which does not use the "res" thing used by Random Variable. The challenge is still open. :)

It's called a residue. Someday... ;)

 

[alyafey22]

Spoiler

$$I = Re \int^\pi_0 e^{e^{ix}}\,dx = Re \int^\pi_0 \sum_{n\geq 0} \frac{e^{inx}}{n!} \, dx=\int^\pi_0 \sum_{n\geq 0} \frac{\cos(nx)}{n!}=\int^\pi_0 1 \, dx +\int^\pi_0 \sum_{n\geq 1}\frac{\cos(nx)}{n!} \,dx=\pi+0 =\pi$$

 

[Saitama]

Spoiler

$$I = Re \int^\pi_0 e^{e^{ix}}\,dx = Re \int^\pi_0 \sum_{n\geq 0} \frac{e^{inx}}{n!} \, dx=\int^\pi_0 \sum_{n\geq 0} \frac{\cos(nx)}{n!}=\int^\pi_0 1 \, dx +\int^\pi_0 \sum_{n\geq 1}\frac{\cos(nx)}{n!} \,dx=\pi+0 =\pi$$

Excellent!