Evaluate the integral ∫−∞∞cos⁡(x)x2+1dx

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Thomas White · Accepted Answer

Method 1:Note that the function eizz2+1 has poles at ±i. Then, by Cauchy&#039;s Integral Formula we have for R&amp;gt;1∮CReizz2+1dz=∫−RReixx2+1dx+∫eiReiϕ(Reiϕ)1+1iReiϕdϕ (1)=2πiei(i)2i=πeAs 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∫0∞cos⁡(x)x2+1dx=π2eMethod 2:Let f(a) be given by the convergent improper integralf(a)=∫0∞cos⁡(ax)x2+1dx (3)Since the integral ∫0∞xsin⁡(ax)x2+1dx is uniformly convergent for |a|≥δ&amp;gt;0, we may differentiate under the integral in (3) for |a|&amp;gt;δ&amp;gt;0 to obtainf′(a)=−∞0∞xsin⁡(ax)x2+1dx=−∫0∞(x2+1−1)sin⁡(ax)x(x2+1)dx=−∫0∞sin⁡(ax)xdx+∫0∞sin⁡(ax)x(x2+1)dx=−π2+∫0∞sin⁡(ax)x(x2+1)dxAgain, since the integ

Alex Sheppard · Accepted Answer

We may also see that  I=∫−∞∞cos⁡(x)1+x2dx=2∫0∞cos⁡(x)1+x2dx  =∫0∞eix+e−ix1+x2dx=e−12(∫0∞e1+ix+e1−ix1+ixdx+∫0∞e1+ix+e1−ix1−ixdx)  =e−12i(∫0∞1x(ixe1+ix1+ix+ixe1−ix1−ix)dx+∫0∞1x(ixe1−ix1+ix+ixe1+ix1−ix)dx)  and now applying the complex version of Frullanis

karton · Accepted Answer

Another approach: A combination of Feynman&#039;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. &#039;t&#039;: L[I(t)]=∫−∞∞L[cos⁡(xt)]x2+1dx=∫−∞∞ss2+x2⋅1x2+1dx=ss2−1∫−∞∞[1x2+1−1x2+s2]dx=ss2−1[arctan⁡(x)−1sarctan⁡(xs)]−∞∞=ss2−1[(π2−1s⋅π2)−(−π2−1s⋅−π2)]=ss2−1[π−π1s]=ss2−1⋅s−1sπ=πs+1 We now take the inverse Laplace Transform: I(t)=L−1[πs+1]=πe−t And finally: J=I(1)=πe−1=πe Now this is essentially residue analysis, but I&#039;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

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