INTEGRAL REPRESENTATION EULER-MASCHERONI CONSTANT

          Today we will proof the following integral representation of the Euler Mascheroni constant that appears in this post



\int_{0}^{\infty }\frac{\cos x^2-\cos x}{x}dx=\frac{\gamma}{2}


We will first proof the following Lemma

Lemma:


\int_{0}^{\infty} \frac{\cos x-e^{-x}}{x} d x=0


Proof: Consider the following complex function

f(z)= \frac{e^{iz}}{z}

and lets integrate it around the contour C below.



Clearly from Cauchy´s theorem the integral is equal to zero

\oint f(z) d z=0

Now,


\begin{aligned} 0&=\int_{\epsilon}^{R} \frac{e^{i x}}{x} d x+\int_{\epsilon}^{R} \frac{e^{i R e^{i \theta}}}{R e^{i \theta}} i R e^{i \theta} d \theta+\int_{R}^{\epsilon} \frac{e^{i(i x)}}{x} i d x+\int_{\pi / 2}^{0} \frac{e^{i \epsilon e^{i \theta}}}{\epsilon e^{i \theta}} i \epsilon e^{i \theta} d \theta\\ &=\int_{\epsilon}^{R} \frac{e^{i x}-e^{-x}}{x} d x+i \int_{0}^{\pi / 2}\left(e^{i R \cdot \cos \theta-R \sin \theta-e^{i c \cos \theta-c \sin \theta}}\right) d \theta\\ \end{aligned}


Taking limits R \rightarrow \infty \,\,\text{and}\,\,\epsilon \rightarrow 0


\begin{gathered} \int_{0}^{\infty} \frac{\cos x+i \sin x-e^{-x}}{x} d x+i \int_{0}^{\pi / 2}(-1) d \theta=0 \\ \int_{0}^{\infty} \frac{\cos x-e^{-x}}{x} d x+i \int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{i \pi}{2} \end{gathered}


Equating Real and Imaginary parts we obtain


\begin{gathered} \int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2} \\ \int_{0}^{\infty} \frac{\cos x-e^{-x}}{x} d x=0 \qquad \blacksquare\\ \end{gathered}(1)


From (1) we conclude that

\int_0^\infty \frac{\cos x}{x}dx=\int_0^\infty \frac{e^{-x}}{x}dx(2)


Letting x \mapsto x^2 in (1), then


2\int_0^\infty \frac{\cos x^2-e^{-x^2}}{x}dx=0

\int_0^\infty \frac{\cos x^2}{x}dx=\int_0^\infty \frac{e^{-x^2}}{x}dx(3)

From (2) and (3) it follows that


\int_{0}^{\infty }\frac{\cos x-\cos x^2}{x}dx=\int_{0}^{\infty }\frac{e^{-x} -e^{ -x^2}}{x}dx(4)


Lets now compute the R.H.S. of (4)


\begin{aligned} \int_{0}^{\infty }\frac{e^{-x} -e^{ -x^2}}{x}dx&=\int_{0}^{\infty }\frac{e^{-x} }{x}dx-\int_0^\infty \frac{e^{ -x^2}}{x}dx \qquad (\text{integrating by parts})\\ &=\int_0^\infty e^{-x} \ln x \,dx-2\int_0^\infty e^{-x^2} x\ln x \,dx \qquad (\text{let} x^2 \mapsto x \,\, \text{in the second integral})\\ &=\int_0^\infty e^{-x} \ln x \,dx+\frac{1}{2}\int_0^\infty e^{-x} \ln x \,dx\\ &=-\frac{\gamma}{2} \qquad \blacksquare \end{aligned}


Where in the last line we used the result proved here. Therefore we conclude that


\int_{0}^{\infty }\frac{\cos x-\cos x^2}{x}dx=-\frac{\gamma}{2}


Or

\boxed{\int_{0}^{\infty }\frac{\cos x^2-\cos x}{x}dx=\frac{\gamma}{2}}

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