∫0∞cos(xt)1+t2dtI'm supposed to solve this using Laplace Transformations.

Let f(x) denote the integral, and assume temporarily that x&gt;0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on s∈(0,∞) (in fact, it defines an analytic function on R(s)&gt;0). Thus we may assume further that s≠1 and then rely on the continuity argument. ThenLf(s)=∫0∞f(x)e−sxdx=∫0∞∫0∞cos⁡(xt)1+t2e−sxdtdx=⋅∫0∞∫0∞cos⁡(xt)1+t2e−sxdxdt=∫0∞11+t2ss2+t2dt=s1−s2∫0∞(1s2+t2−11+t2)dt=s1−s2π2(1s−1)=π211+sHere, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for s≠1, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find thatf(x)=π2e−x

Resolved: "∫0∞cos(xt)1+t2dtI'm supposed to solve this using Laplace Transformations."

Roy Roth

Answered question

2022-03-31

$\int_{0}^{\infty} \frac{\cos (x t)}{1 + t^{2}} d t$
I'm supposed to solve this using Laplace Transformations.

Answer & Explanation

Aarlenlsi1

Beginner2022-04-01Added 10 answers

Let f(x) denote the integral, and assume temporarily that x>0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on

s \in (0, \infty)

(in fact, it defines an analytic function on R(s)>0). Thus we may assume further that

s \neq 1

and then rely on the continuity argument. Then

L f (s) = \int_{0}^{\infty} f (x) e^{- s x} d x

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

\overset{\cdot}{=} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\cos (x t)}{1 + t^{2}} e^{- s x} d x d t

= \int_{0}^{\infty} \frac{1}{1 + t^{2}} \frac{s}{s^{2} + t^{2}} d t

= \frac{s}{1 - s^{2}} \int_{0}^{\infty} (\frac{1}{s^{2} + t^{2}} - \frac{1}{1 + t^{2}}) d t

= \frac{s}{1 - s^{2}} \frac{π}{2} (\frac{1}{s} - 1) = \frac{π}{2} \frac{1}{1 + s}

Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for

s \neq 1

, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that

f (x) = \frac{π}{2} e^{- x}

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