A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:∞∫0sinxxdx=π2

Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of sinx:∞∫0sinxxdx=limn→∞n∫01t(t−t33!+t55!+⋯)dtbut I don’t see how π comes here, since we need the answer to be equal to π2.

In the book

Advanced Calculusby Angus Taylor it is shown that, if a>0a>0,∫∞0e−atsinxttdt=arctanxa.(1)(1)∫0∞e−atsinxttdt=arctanxa.

If x>0x>0,

∫∞0sinxttdt=π2(2)(2)∫0∞sinxttdt=π2

follows from (1)(1), observing that the integrand is G(0)G(0) for

G(a)=∫∞0e−atsinxttdt,(3)(3)G(a)=∫0∞e−atsinxttdt,

GG is uniformly convergent when a≥0a≥0, and G(a)G(a) approaches G(0)G(0) as aa tends to 0+0+

I would like to present yet another simple proof that goes through Fourier series. However, we will need the following theorem; we denote by Sn(x;f)Sn(x;f) the nn-th partial sum of the Fourier series of f(∈L1[−π,π]f(∈L1[−π,π] and 2π2π-periodic) at xx. Then:

Theorem.Riemann’s principle of localization. If f∈L1[−π,π]f∈L1[−π,π], thenSn(x,f)=1π∫δ−δf(x+t)sinnttdt+o(1).(δ>0)Sn(x,f)=1π∫−δδf(x+t)sinnttdt+o(1).(δ>0)

Now, if we pick the function f(x)≡1f(x)≡1, then Sn(x,f)≡1Sn(x,f)≡1 for all n,xn,x. Thus, by Riemann’s principle of localization:

1=1π∫δ−δsinnttdt+o(1)=2π∫nδ0sinttdt+o(1).1=1π∫−δδsinnttdt+o(1)=2π∫0nδsinttdt+o(1).

Letting n→∞n→∞, we get

1=2π∫∞0sinttdt,1=2π∫0∞sinttdt,

which yields the desired result.

Let

I=∫∞0sinxxdx.I=∫0∞sinxxdx.

By the Schwinger parametrization we get

I=∫∞0dt∫∞0sinxexp(−tx)dx.I=∫0∞dt∫0∞sinxexp(−tx)dx.

The last integral can be evaluated by parts. Another simple way is using sinx=I[e−ix]sinx=ℑ[e−ix]:

∫∞0sinxexp(−tx)dx=I∫∞0e−(t−i)xdx=11+t2.∫0∞sinxexp(−tx)dx=ℑ∫0∞e−(t−i)xdx=11+t2.

Thereby,

I=∫∞0dt1+t2.I=∫0∞dt1+t2.

Here you can use again the Schwinger trick. However,

darctan(x)dx=11+x2,darctan(x)dx=11+x2,

Shuch that