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Calculus Level 5

Find $\displaystyle \int _{ 0 }^{ \infty }{ \frac { \cos { \left( tx \right) } }{ { x }^{ 2 }+1 } \,dx }$ at $t=\ln { \left( \dfrac { \pi }{ 8 } \right) }$ .

If this integral can be expressed as $A\times \pi^B$ for rationals $A$ and $B$ , find $A\times B$ .

The answer is 0.125.

3 solutions

Rishabh Deep Singh
Feb 13, 2017

Yes differentiation under the integral sign makes a clear cut solution.Your approach is exactly the same.For the general case of $\int cos(bx)/(x^2+a^2)$ = $π/2e^(-ab)$ with the same limits.

Spandan Senapati - 4 years, 1 month ago

I am getting the answer as pi^2/16

Vijay Simha - 4 years, 1 month ago

No, my friend, I think you miss the negative sign of exponential.

Rishabh Deep Singh - 4 years, 1 month ago

$\frac{\pi^2}{16}$ is the correct answer. You are missing an absolute value in your solution- $f(t) = \frac{\pi}{2} e^{-|t|}$ .

D G - 3 years, 8 months ago

@D G – Specifically in the step from line 4 to 5, taking the limit as x -> infinity gives $(\pi - \pi \sqrt{(1/s^2)}) / 2 = \frac{\pi}{2} - \frac{\pi}{2 |s|}$

D G - 3 years, 8 months ago

Thanks. I've updated the problem statement to reflect this.

In the future, if you have concerns about a problem's wording/clarity/etc., you can report the problem. See how here .

Brilliant Mathematics Staff - 3 years, 8 months ago

Can you post a link where I can learn this method?

Akshat Sharda - 3 years, 8 months ago

Its a pretty standard calculus,question.If you know Feynman's Trick,then you can solve it.Although that method also uses some knowledge of second order differential equations.Refer to any good integral book,like 'Irrestible Integrals'by Victor,Moll.

Spandan Senapati - 3 years, 8 months ago

What is 's' here?

Akshat Sharda - 3 years, 8 months ago

We need the Staff to resolve this Is the answer pi^2/16 or something else.

Vijay Simha - 3 years, 8 months ago

@Rishabh Deep Singh - What is the script L with the curly brackets in your solution? Also, where did s come from?

Christopher Criscitiello - 3 years, 8 months ago

That is notation for the Laplace transform.

D G - 3 years, 8 months ago

@Christopher Criscitiello That comes from the Laplace transformation.

Rishabh Deep Singh - 11 months, 2 weeks ago

Chew-Seong Cheong
Sep 19, 2017

Using contour integration as suggested by @Mr. Math .

$\begin{aligned} I & = \int_0^\infty \frac {\cos (tx)}{x^2+1} dx &\small \color{#3D99F6} \text{Since the integrand is even,} \\ & = \frac 12 \int_{-\infty}^\infty \frac {\cos (tx)}{x^2+1} dx & \small \color{#3D99F6} C\text{: infinite semicircle in upper half-plain} \\ & = \frac 12 \int_C \frac {e^{itz}}{z^2+1} dz & \small \color{#3D99F6} \text{Integral around the arc}=0 \\ & = \frac 12 \int_C \frac {e^{itz}}{(z+i)(z-i)} dz & \small \color{#3D99F6} \text{The only pole in contour is }z = i \\ & = \frac 12 \cdot 2 \pi i \cdot \frac {e^{-t}}{2i} & \small \color{#3D99F6} \text{By residue theorem} \\ & = \frac {\pi e^{-\ln \frac \pi 8}}2 \\ & = \frac \pi{2\cdot \frac \pi 8} \\ & = \boxed{4} \end{aligned}$

Unfortunately, you cannot take the infinite semicircle in the upper half plane. If you consider $z \rightarrow ni$ , then $\frac{ e^{itz} } { z^2 + 1 } = \frac {e^{-n t }} { -n^2 +1 }$ . However, note that because $t$ is negative, this term goes to infinity.

As such, we have to take the infinite semicircle in the lower half plane , and thus the answer is $\pi e^{|t| }$ .

Calvin Lin Staff - 3 years, 7 months ago

Mr. Math
Mar 4, 2017

I solved it using contour integration

Integration is back

The answer is 0.125.

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