Real Analysis Methodologies to Prove $-\int_x^\infty \frac{\cos(t)}{t}\,dt=\gamma+\log(x)+\int_0^x \frac{\cos(t)-1}{t}\,dt$ for $x>0$

Question

MOTIVATION:

In this question, the OP asked to prove the Cosine Integral identity

$$-\int_x^\infty \frac{\cos(t)}{t}\,dt=\gamma+\log(x)+\int_0^x \frac{\cos(t)-1}{t}\,dt$$

where $\gamma $ is the Euler-Mascheroni constant. However, the two posted answers are incomplete and seem unsatisfactory.

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Proof Using Complex Anaysis

One can show for $x>0$ that $-\int_x^\infty \frac{\cos(t)}{t}\,dt=\gamma+\log(x)+\int_0^x \frac{\cos(t)-1}{t}\,dt$ using contour integration. To wit, Cauchy's Integral Theorem guarantees that

$$\begin{align} 0&=\oint_C \frac{e^{iz}}{z}\,dz\\\\ &=\int_\epsilon^R \frac{e^{ix}}{x}\,dx +\int_0^{\pi/2}\frac{e^{iRe^{i\phi}}}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi+\int_R^\epsilon \frac{e^{-x}}{ix}\,i\,dx+\int_{\pi/2}^0 \frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\\\\ &=\int_\epsilon^R \frac{e^{ix}}{x}\,dx -\int_\epsilon^R \frac{e^{-x}}{x}\,dx-i\frac\pi2+O(\epsilon)+O\left(\frac1R\right)\tag1 \end{align}$$

whence after taking the real part of both sides of $(1$ and integrating by parts the integral $\int_\epsilon^R \frac{e^{-x}}{x}\,dx$ with $u=e^{-x}$ and $v=\log(x)$, we find

$$\begin{align}-\int_x^R \frac{\cos(x')}{x'}\,dx'&=\log(x)-\int_\epsilon^R e^{-x}\log(x)\,dx+\int_\epsilon^x\frac{\cos(x')-1}{x'}\,dx'\\\\ &-\log(\epsilon) -e^{-R}\log(R)+e^{-\epsilon}\log(\epsilon)+O\left(\frac1R\right)+O(\epsilon)\tag2 \end{align}$$

Letting $R\to\infty$ and $\epsilon\to 0$ in $(2)$ yields the sought relationship.

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Proof Using Real Analysis

Alternatively, we can use either the Laplace Transform or "Feynman's Trick" to show that

$$\int_0^\infty \frac{\cos(x)-e^{-x}}{x}\,dx=\int_0^\infty \left(\frac{x}{x^2+1}-\frac1{x+1}\right)=0\tag3$$

Starting with $(3)$, is tantamount to starting with the real part of $(1)$ and we are done.

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QUESTION: So, what are other ways to prove the coveted relationship using real analysis tools only?

Answer 1

Note that for each $x>0$, the improper integral $\int_{x}^{\infty}\frac{\cos(t)}{t}dt$ converges conditionally (in the sense that $\lim_{A\rightarrow\infty}\int_{x}^{A}\frac{\cos(t)}{t}dt$ exists but $\lim_{A\rightarrow\infty}\int_{x}^{A}\left|\frac{\cos(t)}{t}\right|dt=\infty$). Define $F:(0,\infty)\rightarrow\mathbb{R}$ by $F(x)=\int_{x}^{\infty}\frac{\cos(t)}{t}dt$. Note that we still have $F'(x)=-\frac{\cos(x)}{x}$. For, \begin{eqnarray*} F'(x) & = & \lim_{h\rightarrow0}\frac{\int_{x+h}^{\infty}\frac{\cos(t)}{t}dt-\int_{x}^{\infty}\frac{\cos(t)}{t}dt}{h}\\ & = & -\lim_{h\rightarrow0}\frac{\int_{x}^{x+h}\frac{\cos(t)}{t}dt}{h}\\ & = & -\frac{\cos(x)}{x}. \end{eqnarray*}

On the other hand, $0$ in the integral $\int_{0}^{x}\frac{\cos(t)-1}{t}dt$ is a removable singularity. For $t\neq0$, we have \begin{eqnarray*} \frac{\cos(t)-1}{t} & = & \frac{(1-\frac{1}{2!}t^{2}+\frac{1}{4!}t^{4}-\cdots)-1}{t}\\ & = & -\frac{1}{2!}t+\frac{1}{4!}t^{3}-\frac{1}{6!}t^{5}+\cdots. \end{eqnarray*} Define $\phi:\mathbb{R}\rightarrow\mathbb{R}$ by $\phi(t)=-\frac{1}{2!}t+\frac{1}{4!}t^{3}-\frac{1}{6!}t^{5}+\cdots$. It can be proved that the power series converges everywhere (using root-test), and hence $\phi$ is an analytic function. Therefore $\int_{0}^{x}\frac{\cos(t)-1}{t}dt=\int_{0}^{x}\phi(t)dt$.

Now $\frac{d}{dx}\int_{0}^{x}\frac{\cos(t)-1}{t}dt=\frac{\cos(x)-1}{x}$.

Finally, define $G:(0,\infty)\rightarrow\mathbb{R}$ by $$ G(x)=\int_{x}^{\infty}\frac{\cos(t)}{t}dt-\left[\ln(x)+\int_{0}^{x}\frac{\cos(t)-1}{t}dt\right]. $$ Then $G$ is differentiable and $G'(x)=-\frac{\cos(x)}{x}-\frac{1}{x}-\frac{\cos(x)-1}{x}=0$. This shows that $G$ is a constant function.