How do you find the integral of #int dx/(x^2+2)^(3/2)# from negative infinity to infinity?

Question

Isabella Ackerman · Accepted Answer

#int_(-oo)^oo1/(x^2+2)^(3/2)dx = 1#

First, let's solve the indefinite integral with trig substitution:

#int1/(x^2+2)^(3/2)dx = int1/(2sqrt(2)((x/sqrt(2))^2+1)^(3/2))dx#

Let #x/sqrt(2) = tan(theta) => 1/sqrt(2)dx = sec^2(theta)d theta#

#=>int1/(2sqrt(2)((x/sqrt(2))^2+1)^(3/2))dx = 1/2intsec^2(theta)/(tan^2(theta)+1)^(3/2)d theta#

#=1/2intsec^2(theta)/sec^3(theta)d theta#

#=1/2intcos(theta)d theta#

#=1/2sin(theta) + C#

#=x/(2sqrt(x^2+2)) + C#

(To do the last step, try drawing the right triangle with an angle #theta# such that #tan(theta) = x/sqrt(2)# and then evaluate #sin(theta)#)

Now, we can look at the definite integral from #-oo# to #oo#

#int_(-oo)^(oo)1/(x^2+2)^(3/2) = lim_(A->-oo)lim_(B->oo)int_A^B1/(x^2+2)^(3/2)dx#

#=lim_(A->-oo)lim_(B->oo)[x/(2sqrt(x^2+2))]_A^B#

#=lim_(A->-oo)lim_(B->oo)1/2(B/sqrt(B^2+2)-A/sqrt(A^2+2))#

#=lim_(A->-oo)lim_(B->oo)1/2(1/(1/Bsqrt(B^2+2))-1/(1/Asqrt(A^2+2)))#

#=lim_(A->-oo)lim_(B->oo)1/2(1/sqrt((B^2+2)/B^2)+1/sqrt((A^2+2)/A^2))#

(We actually cheated a little bit on this step, as we implicitly used that #1/B = 1/sqrt(B^2)# and #1/A = -1/sqrt(A^2)#, which is true given that #B# will be positive and #A# will be negative as we take the limit, but isn't strictly valid to say at this step. It does, however, save us some messy algebra/calculus, and we get the same result)

#= lim_(A->-oo)lim_(B->oo)1/2(1/sqrt(1+2/B^2) + 1/sqrt(1+2/A^2))#

#= lim_(A->-oo)1/2(1 + 1/sqrt(1+2/A^2))#

#=1/2(1+1)#

#=1#

Samantha Adkison · Answer

undefined To find the integral of $ \int \frac{dx}{(x^2+2)^{3/2}} $ from negative infinity to infinity, you can use the method of residues from complex analysis. The integral can be represented as the contour integral of a complex function along a semicircular contour in the upper half-plane, which closes either above or below, depending on the sign of the exponent. Using the residue theorem, you can then find the residues of the function within the contour and sum them up. The integral over the semicircular arc tends to zero as the radius of the semicircle approaches infinity, leaving only the contribution from the residues. The result of the integral is then determined by the residues of the function at its singularities.