Solved: "Calculate the improper integral ∫0∞1θecos⁡θsin⁡(sin⁡θ)dθ"

Kelly Nelson

Answered question

2022-01-14

Calculate the improper integral

\int_{0}^{\infty} \frac{1}{θ} e^{\cos θ} \sin (\sin θ) d θ

Answer & Explanation

Stuart Rountree

Beginner2022-01-15Added 29 answers

Step 1
Define the sine integral by
$S i (x) = \int_{0}^{x} \frac{\sin t}{t} d t$
Using integartion by parts, it can be proved that
$S i (x) = \frac{π}{2} + O (\frac{1}{x})$ as $x \to \infty$
Now note that
$e^{\cos θ} \sin \sin θ = I m (e^{e^{1 θ}} - 1) = \sum_{n = 1}^{\infty} \frac{1}{n!} \sin (n θ)$
Then by the Fubini's theorem, for $R > 0$
$\int_{0}^{R} \frac{e^{\cos θ} \sin \sin θ}{θ} d θ = \int_{0}^{R} \frac{1}{θ} \sum_{n = 1}^{\infty} \frac{\sin (n θ)}{n!} d θ$
$= \sum_{n = 1}^{\infty} \frac{1}{n!} \int_{0}^{R} \frac{\sin (n θ)}{θ} d θ$
$= \sum_{n = 1}^{\infty} \frac{1}{n!} S i (n R)$
$= \sum_{n = 1}^{\infty} \frac{1}{n!} (\frac{π}{2} + O (\frac{1}{n R}))$
$= \frac{π}{2} (e - 1) + O (\frac{1}{R})$
So by letting $R \to \infty$ , the integral converges to
$\int_{0}^{\infty} \frac{e^{\cos θ} \sin \sin θ}{θ} d θ = \frac{π}{2} (e - 1)$

Juan Spiller

Beginner2022-01-16Added 38 answers

Step 1
It is well-known that

lim_{N \to \infty} \sum_{k = - N}^{N} \frac{1}{z + 2 π k} = \frac{1}{2} \cot (\frac{z}{2})

Moreover, this convergence is locally uniform (in the sense that the difference between the limit and the N-th partial sum, when understood as a meromorphic function on

C

, converges to 0 uniformly on any compact subsets of

C

).
Using this and noting that

e^{\cos θ} \sin \sin θ = I m (e^{e^{1 θ}} - e)

, we find

θ = I m (\int_{- (2 N + 1) π}^{(2 N + 1) π} \frac{e^{e^{1 θ}}}{θ} d θ)

= I m (\int_{- π}^{π} (e^{e^{1 θ}} - e) \sum_{k = - N}^{N} \frac{1}{θ + 2 π k} d θ)

\to I m (\int_{- π}^{π} (e^{e^{1 θ}} - e) \frac{1}{2} \cot (\frac{θ}{2}) d θ)

N \to \infty

Now we substitute

z = e^{i θ}

. Then using the identity

\cot (\frac{θ}{2}) = i \frac{e^{i θ} + 1}{e^{i θ} - 1}

and the residue theorem,

\int_{- \infty}^{\infty} \frac{e^{\cos θ} \sin \sin θ}{θ} d θ = I m (\frac{1}{2} \int_{| z | = 1} \frac{(e^{z} - e) (z + 1)}{z (z - 1)} d z)

= I m (π i R e s_{z = 0} \frac{(e^{z} - e) (z + 1)}{z (z - 1)})

= π (e - 1)

Dividing both sides by 2, we conclude that

\int_{0}^{\infty} \frac{e^{\cos θ} \sin \sin θ}{θ} d θ = \frac{π}{2} (e - 1)

alenahelenash

Expert2022-01-24Added 556 answers

Step 1 Its

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