Integration of sinx/x from 0 to infinity?

1 Answer
Sep 5, 2017

# int_0^oo \ sinx/x \ dx = pi/2#

Explanation:

We seek:

# I = int_0^oo \ sinx/x \ dx #

Let #g(x) = sinx/x => g(-x) = sin(-x)/(-x) = sinx/x #

Thus #g(x)# is an even function, and as such:

# 2I = int_(-oo)^oo \ sinx/x \ dx #

Consider the complex based function # f(z)=e^(iz)/z #, Which has a simple pole at #z=0#, we then consider the contour integral:

# J = oint_C \ f(z) \ dz = oint_C \ e^(iz)/z \ dz # where #z in CC#

Where #C# is a semicircle of radius #R# centred at the origin that is deformed at the origin with a smaller semicircle of radius #epsilon# to excludes the pole at #z = 0#, and we traverse the contour in an anticlockwise direction.

Steve M

The integrand has no poles in #C# as the pole #z = 0# is excluded in the above construction. Thus, by Cauchy’s Theorem:

# oint_C \ f(z) \ dz = 0#

Now (in shorthand),

# oint_C \ f(z) \ dz = int_(-R)^(epsilon) + int_(gamma_epsilon) + int_(epsilon)^R + int_(Gamma_R) = 0#

We require an estimate for #int_(Gamma_R) \ f(z) \ dz #. Noting that #z=Re^(i theta)# on #Gamma_R#, we have:

# abs(int_(Gamma_R) \ e^(iz)/z \ dz) = abs(int_o^oo \ e^(iRcos theta-R sin theta) / (Re^(i theta)) iRe^(i theta) \ d theta ) #
# " " le int_0^pi e^(-Rsin theta) \ d theta #
# " " = 2 int_0^(pi/2) e^(-Rsin theta) \ d theta #
# " " le 2 int_0^(pi/2) e^((-2R theta) / pi) \ d theta \ \ \ \# using #sin theta ge (2theta)/pi#
# " " = 2 [ (2e^((-2Rtheta)/pi) )/ ((-2R)/pi) ]_0^(pi/2)#
# " " = pi/R(1-e^-R)#
# " " rarr 0 # as #R rarr oo#

Given that the small circle #gamma_epsilon# has the equation #z = r(cos theta + isin theta)# for #theta:pi rarr 0# then

# lim_(epsilon rarr 0) int_(gamma_epsilon) \ f(z) \ dz = i \ lim_(epsilon rarr 0) int_pi^0 \ e^(-rsin theta)e^(ircos theta) \ dz #
# " " = -pi i #

Taking the two limits #R rarr oo# and #epsilon rarr 0#, and combining all these results, we have:

# int_(-oo)^oo e^(iz)/z \ dz - pi i = 0 => int_(-oo)^oo (cosx+isinx)/x \ dx = pi i #

Equating real and imaginary coefficients we get:

# Re: int_(-oo)^oo cosx/x \ dx = 0 #
# Im: int_(-oo)^oo sinx/x \ dx = pi #

Then using the initial result

# 2I = int_(-oo)^oo \ sinx/x \ dx => 2I = pi#

Hence,

# I = pi/2 #