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

Sapewa 2022-01-05 Answered
Evaluate the integral cos(x)x2+1dx
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Thomas White
Answered 2022-01-06 Author has 40 answers

Method 1:
Note that the function eizz2+1 has poles at ±i. Then, by Cauchy's Integral Formula we have for R>1
CReizz2+1dz=RReixx2+1dx+eiReiϕ(Reiϕ)1+1iReiϕdϕ (1)
=2πiei(i)2i
=πe
As R, the second integral on the right-hand side of (1) approaches 0. Therefore, we find that
eixx2+1dx=πe (2)
Tacking the real part of both sides of (2) and exploiting the even symmetry yields
0cos(x)x2+1dx=π2e
Method 2:
Let f(a) be given by the convergent improper integral
f(a)=0cos(ax)x2+1dx (3)
Since the integral 0xsin(ax)x2+1dx is uniformly convergent for |a|δ>0, we may differentiate under the integral in (3) for |a|>δ>0 to obtain
f(a)=0xsin(ax)x2+1dx
=0(x2+11)sin(ax)x(x2+1)dx
=0sin(ax)xdx+0sin(ax)x(x2+1)dx
=π2+0sin(ax)x(x2+1)dx
Again, since the integ

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Alex Sheppard
Answered 2022-01-07 Author has 36 answers
We may also see that
I=cos(x)1+x2dx=20cos(x)1+x2dx
=0eix+eix1+x2dx=e12(0e1+ix+e1ix1+ixdx+0e1+ix+e1ix1ixdx)
=e12i(01x(ixe1+ix1+ix+ixe1ix1ix)dx+01x(ixe1ix1+ix+ixe1+ix1ix)dx)
and now applying the complex version of Frullanis
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karton
Answered 2022-01-11 Author has 439 answers

Another approach: A combination of Feynman's Trick and Laplace Transforms:
J=cos(x)x2+1dx
Here let:
I(t)=cos(xt)x2+1dx
We see I(1)=J and I(0)=π. Here we take the Laplace Transform w.r.t. 't':
L[I(t)]=L[cos(xt)]x2+1dx=ss2+x21x2+1dx=ss21[1x2+11x2+s2]dx=ss21[arctan(x)1sarctan(xs)]=ss21[(π21sπ2)(π21sπ2)]=ss21[ππ1s]=ss21s1sπ=πs+1
We now take the inverse Laplace Transform:
I(t)=L1[πs+1]=πet
And finally:
J=I(1)=πe1=πe
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.

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