# Question #3675d

Nov 21, 2017

This integral converged to $\frac{\pi}{4}$

#### Explanation:

$I = {\int}_{0}^{\infty} \cos \frac{2 t}{t} \cdot \mathrm{dt}$

I use Feynman's trick for evaluating integral,

$I \left(a\right) = {\int}_{0}^{\infty} \frac{{e}^{- a t} \cdot \cos \left(2 t\right)}{t} \cdot \mathrm{dt}$

I note that original integral can be reached after setting $b = 0$.

After differentiating under integral sign according to $a$,

$I ' \left(a\right) = {\int}_{0}^{\infty} - {e}^{- a t} \cdot \cos 2 t \cdot \mathrm{dt}$

=$\frac{1}{a} \cdot {\left[{e}^{- a t} \cdot \cos 2 t\right]}_{0}^{\infty} + \frac{2}{a} \cdot {\int}_{0}^{\infty} {e}^{- a t} \cdot \sin 2 t \cdot \mathrm{dt}$

=$- \frac{1}{a} + \frac{2}{a} \cdot {\int}_{0}^{\infty} {e}^{- a t} \cdot \sin 2 t \cdot \mathrm{dt}$

=$- \frac{1}{a} - \frac{2}{a} ^ 2 \cdot {\left[{e}^{- a t} \cdot \sin 2 t\right]}_{0}^{\infty}$-$\frac{4}{a} ^ 2 \cdot {\int}_{0}^{\infty} {e}^{- a t} \cdot \cos 2 t \cdot \mathrm{dt}$

=$- \frac{1}{a} - \frac{2}{a} ^ 2 \cdot 0 - \frac{4}{a} ^ 2 \cdot I ' \left(a\right)$

$\left(1 + \frac{4}{a} ^ 2\right) \cdot I ' \left(a\right) = - \frac{1}{a}$

$I ' \left(a\right) = - \frac{a}{{a}^{2} + 4}$

After integrating both sides,

$I \left(a\right) = - \frac{1}{2} \cdot \arctan \left(\frac{a}{2}\right) + C$

Now, I choose a strategic value $b = {b}_{0}$ in order to make our integrand vanish so that $I \left({b}_{0}\right) = 0$. In this case, took $b = \infty$, so that $I \left(\infty\right) = 0$

Hence $- \frac{1}{2} \cdot \arctan \left(\frac{\infty}{2}\right) + C = 0$, so $C = \frac{\pi}{4}$

Thus $I = I \left(0\right) = - \frac{1}{2} \cdot \arctan \left(\frac{0}{2}\right) + \frac{\pi}{4} = \frac{\pi}{4}$