Integral of Dirichlet

The integral of Dirichlet is the integral of the function cardinal Sinus on the half-line of positive realities

\ int_0^ {+ \ infty} \ frac {\ sin (X)}{X} \, dx = \ frac {\ pi} {2}

It is about a convergent Intégrale unsuitable, i.e. the function is not integrable with the generalized direction of Riemann, but $\ lim_ {X \ to + \ infty} \ int_0^ {X} \ frac {\ sin (X)}{X} \, dx$ exists.

Proof

One considers the function: $\ begin {matrix} F: & \ mathbb {R} _ {+} ^ {*} & \ rightarrow & \ mathbb {R} \ \ & X & \ mapsto & \ frac {\ sin (X)}{X} \ \ \ \ end {matrix}$

There is $\ lim_ {0^ {+}} F = 1 \$ , therefore $F$ is prolongeable by continuity into 0.

Let us show that this function is not integrable: one considers, for all $N \ in \ mathbb {NR}$ , the continuation: $u_n = \ int_ {N \ pi} ^ {(n+1) \ pi} |F (X)|\, dx = \ int_ {N \ pi} ^ {(n+1) \ pi} \ frac {X} dx$ .

Change of variables $T = X - N \ pi$ gives $u_n = \ int_ {0} ^ {\ pi} \ frac {T + N \ pi} dt$ .

One can then write: $u_n \ geq \ frac {1} {(n+1) \ pi} \ int_ {0} ^ {\ pi} \ sin (T) dt = \ frac {2} {(n+1) \ pi}$ .

One from of deduced: $\ forall NR \ in \ mathbb {NR}, \ int_0^ {NR \ pi} \ frac {X} \, dx = \ sum_ {n=0} ^ {N-1} u_n \ geq \ frac {2} {\ pi} \ sum_ {n=1} ^N \ frac {1} {N}$ , but this series is the harmonic Série, which diverges.

The function $F$ is thus not integrable on $\ mathbb {R} _ {+} ^ {*}$ .

Montrons now that $\ lim_ {X \ to + \ infty} \ int_0^ {X} F (X) dx$ exists.

There is $\ int_0^ {X} F (X) dx = \ int_0^ {X} \ frac {1} {X} \ sin (X) dx$ . An integration by parts, with $U (X) = \ frac {1} {X}, v' (X) = \ sin (X)$ , then by taking $u' (X) = - \ frac {1} {x^2}, v (X) =1- \ cos (X)$ , makes it possible to write, wrongly: $\ int_0^ {X} F (X) dx = \ left \ frac {1 - \ cos (X)}{X} \ right _ {0} ^ {X} + \ int_ {0} ^ {X} \ frac {1 - \ cos (X)}{x^2} dx$ .

The abuse comes owing to the fact that $x \ mapsto \ frac {1 - \ cos (X)}{X}$ is not defined into 0. However, as one has into 0: $1- \ cos (X) \ sim \ frac {x^2} {2}$ , one writes: $\ int_0^ {X} F (X) dx = \ frac {1 - \ cos (X)}{X} + \ int_ {0} ^ {X} \ frac {1 - \ cos (X)}{x^2} dx$ .

Moreover, as one has $\ lim_ {X \ to 0} \ frac {1 - \ cos (X)}{x^2} = \ frac {1} {2}$ and $\ forall X \ geq 0, \ left| \ frac {1 - \ cos (X)}{x^2} \right| \ Leq \ frac {2} {x^2}$ , one from of thus deduced that $\ int_ {0} ^ {+ \ infty} \ frac {1 - \ cos (X)}{x^2} dx$ converges.

With $\ lim_ {X \ to + \ infty} \ frac {1 - \ cos (X)}{X} =0$ , one concludes from it that $\ lim_ {X \ to + \ infty} \ int_0^ {X} F (X) dx$ exists.

Calculation of the integral of Dirichlet

With continuations

Let us pose, for $x \ in \ left] 0; \ frac {\ pi} {2} \ right]$ , the function $g (X) = \ frac {1} {X} - \ frac {1} {\ sin (X)}$ .

As one has into 0: $\ sin (X) \ sim x$ and $\ sin (X) - X \ sim - \ frac {x^3} {6}$ , one has thus, always into 0: $g (X) \ sim - \ frac {X} {6}$ .

The function G is thus continuous on on $\ left] 0; \ frac {\ pi} {2} \ right]$ , and prolongeable by continuity in 0.

One considers now the continuation of integrals $J_n = \ int_ {0} ^ {\ frac {\ pi} {2}} \ frac {\ sin \ left ((2n+1) X \ right)}{\ sin (X)} dx$ .

Like $\ lim_ {X \ to 0} \ frac {\ sin \ left ((2n+1) X \ right)}{\ sin (X)}=2n+1$ , the continuation $\ left (J_n \ right) _ {N \ in \ mathbb {NR}}$ is well defined.

Moreover, let us notice that $\ forall N \ in \ mathbb {NR}, \ sin \ left ((2n+1) X \ right) - \ sin \ left ((2n-1) X \ right) = 2 \ sin (X) \ cos (2nx)$ .

One draws $J_n then from it - J_ {n-1} = 2 \ int_ {0} ^ {\ frac {\ pi} {2}} \ cos (2nx) dx = 0$ .

The continuation $J_n$ is thus constant, and $\ forall N \ in \ mathbb {NR}, J_n = J_0 = \ frac {\ pi} {2}$ .

One considers now the continuation of integrals $K_n = \ int_ {0} ^ {\ frac {\ pi} {2}} \ frac {\ sin \ left ((2n+1) X \ right)}{X} dx$ .

Like $\ lim_ {X \ to 0} \ frac {\ sin \ left ((2n+1) X \ right)}{X} =2n+1$ , the continuation $\ left (K_n \ right) _ {N \ in \ mathbb {NR}}$ is well defined.

The change of variables $T = (2n+1) X$ gives $K_n = \ int_ {0} ^ {(2n+1) \ frac {\ pi} {2}} \ frac {\ sin (T)}{T} dt$ .

One from of deduced then $\ lim_ {N \ to + \ infty} K_n = \ int_ {0} ^ {+ \ infty} \ frac {\ sin (X)}{X} dx$ .

One a: $\ forall N \ in \ mathbb {NR}, K_n - J_n = \ int_ {0} ^ {\ frac {\ pi} {2}} G (X) \ sin \ left ((2n-1) X \ right) dx$ .

It was seen that the function G is continuous on $\ left] 0; \ frac {\ pi} {2} \ right]$ , therefore by the lemma of Riemann-Lebesgue, $\ lim_ {N \ to + \ infty} \ left (K_n - J_n \ right) = 0$ .

One concludes from it: $\ int_0^ {+ \ infty} \ frac {\ sin (X)}{X} \, dx = \ frac {\ pi} {2}$ .

With the Remainder theorem

By noticing that $\ int_0^ {+ \ infty} \ frac {\ sin (X)}{X} \, dx = \ frac {1} {2} \ int_ {- \ infty} ^ {+ \ infty} \ frac {\ sin (X)}{X} \, dx = \ frac {1} {2} Im \ left (\ int_ {- \ infty} ^ {+ \ infty} \ frac {e^ {ix}} {X} \, dx \ right)$ , and by considering the function complexes $z \ mapsto \ frac {e^ {iz}} {Z}$ , the remainder theorem directly gives the desired result.

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