Evaluate the integral \int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}dx

Question

Evaluate the integral

\int_{- \infty}^{\infty} \frac{\cos (x)}{x^{2} + 1} d x

Answer 1

Method 1:
Note that the function $\frac{e^{i z}}{z^{2} + 1}$ has poles at $\pm i$ . Then, by Cauchy's Integral Formula we have for $R > 1$
$\oint_{C_{R}} \frac{e^{i z}}{z^{2} + 1} d z = \int_{- R}^{R} \frac{e^{i x}}{x^{2} + 1} d x + \int \frac{e^{i R e^{i ϕ}}}{(R e^{i ϕ)^{1} + 1}} i R e^{i ϕ} d ϕ$ (1)
$= 2 π i \frac{e^{i (i)}}{2 i}$
$= \frac{π}{e}$
As $R \to \infty$ , the second integral on the right-hand side of (1) approaches 0. Therefore, we find that
$\int_{- \infty}^{\infty} \frac{e^{i x}}{x^{2} + 1} d x = \frac{π}{e}$ (2)
Tacking the real part of both sides of (2) and exploiting the even symmetry yields
$\int_{0}^{\infty} \frac{\cos (x)}{x^{2} + 1} d x = \frac{π}{2 e}$
Method 2:
Let $f (a)$ be given by the convergent improper integral
$f (a) = \int_{0}^{\infty} \frac{\cos (a x)}{x^{2} + 1} d x$ (3)
Since the integral $\int_{0}^{\infty} \frac{x \sin (a x)}{x^{2} + 1} d x$ is uniformly convergent for $| a | \geq δ > 0$ , we may differentiate under the integral in (3) for $| a | > δ > 0$ to obtain
$f^{'} (a) = - \infty_{0}^{\infty} \frac{x \sin (a x)}{x^{2} + 1} d x$
$= - \int_{0}^{\infty} \frac{(x^{2} + 1 - 1) \sin (a x)}{x (x^{2} + 1)} d x$
$= - \int_{0}^{\infty} \frac{\sin (a x)}{x} d x + \int_{0}^{\infty} \frac{\sin (a x)}{x (x^{2} + 1)} d x$
$= - \frac{π}{2} + \int_{0}^{\infty} \frac{\sin (a x)}{x (x^{2} + 1)} d x$
Again, since the integ

Answer 2

We may also see that

I = \int_{- \infty}^{\infty} \frac{\cos (x)}{1 + x^{2}} d x = 2 \int_{0}^{\infty} \frac{\cos (x)}{1 + x^{2}} d x

= \int_{0}^{\infty} \frac{e^{i x} + e^{- i x}}{1 + x^{2}} d x = \frac{e^{- 1}}{2} (\int_{0}^{\infty} \frac{e^{1 + i x} + e^{1 - i x}}{1 + i x} d x + \int_{0}^{\infty} \frac{e^{1 + i x} + e^{1 - i x}}{1 - i x} d x)

= \frac{e^{- 1}}{2 i} (\int_{0}^{\infty} \frac{1}{x} (\frac{i x e^{1 + i x}}{1 + i x} + \frac{i x e^{1 - i x}}{1 - i x}) d x + \int_{0}^{\infty} \frac{1}{x} (\frac{i x e^{1 - i x}}{1 + i x} + \frac{i x e^{1 + i x}}{1 - i x}) d x)

and now applying the complex version of Frullanis

Answer 3

Another approach: A combination of Feynman's Trick and Laplace Transforms:
$J = \int_{- \infty}^{\infty} \frac{\cos (x)}{x^{2} + 1} d x$
Here let:
$I (t) = \int_{- \infty}^{\infty} \frac{\cos (x t)}{x^{2} + 1} d x$
We see $I (1) = J$ and $I (0) = π$ . Here we take the Laplace Transform w.r.t. 't':
$L [I (t)] = \int_{- \infty}^{\infty} \frac{L [\cos (x t)]}{x^{2} + 1} d x = \int_{- \infty}^{\infty} \frac{s}{s^{2} + x^{2}} \cdot \frac{1}{x^{2} + 1} d x = \frac{s}{s^{2} - 1} \int_{- \infty}^{\infty} [\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + s^{2}}] d x = \frac{s}{s^{2} - 1} [\arctan (x) - \frac{1}{s} \arctan (\frac{x}{s})]_{- \infty}^{\infty} = \frac{s}{s^{2} - 1} [(\frac{π}{2} - \frac{1}{s} \cdot \frac{π}{2}) - (- \frac{π}{2} - \frac{1}{s} \cdot - \frac{π}{2})] = \frac{s}{s^{2} - 1} [π - π \frac{1}{s}] = \frac{s}{s^{2} - 1} \cdot \frac{s - 1}{s} π = \frac{π}{s + 1}$
We now take the inverse Laplace Transform:
$I (t) = L^{- 1} [\frac{π}{s + 1}] = π e^{- t}$
And finally:
$J = I (1) = π e^{- 1} = \frac{π}{e}$
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.

Evaluate the integral \int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}dx

Expert Answer

Not exactly what you’re looking for?

Not exactly what you’re looking for?

Not exactly what you’re looking for?

You might be interested in

New questions