Limit with Integration - 3

nice solution

Thanks^^ Just noticed the wiki-page on "Differentiation under the integral sign" lists the first integral as a counter-example (which is true for the wrong parameter choice), so I'll add a rough sketch of how to calculate the "Dirichlet-Integral": $I(t):=\int_0^\infty \frac{\sin(y)}{y}e^{-yt}\:dy=:\int_0^{\infty} f(y,\:t)\:dy,\qquad \lim_{t\rightarrow\infty}I(t)=0,\qquad I(0)=??$ For $t>0$ , one can use differentiation under the integral sign: $I^{(1)}(t)=-\int_0^\infty \sin(y)e^{-yt}\:dy=-\Im\left\{\int_0^\infty e^{(i-t)y}\:dy\right\}=-\Im\left\{\frac{-1}{i-t}\right\}=-\frac{1}{1+t^2}$ The remaining integral is simple, and the boundary condition at $t\rightarrow+\infty$ determines the integration constant: $I(t)=-\arctg(t)+\frac{\pi}{2}\quad\Rightarrow\quad I(0)=\frac{\pi}{2}$

Rem.: The above is just a rough sketch - there are (at least) two details missing:

• Why is $I(t)$ continuous from above at $t=0$ ?
• Why are we allowed to use differentiation under the integral sign for an unbounded integration domain?

Suppose that $|h| < \tfrac12t$ , where $t > 0$ , then $\left| \frac{e^{-y(t+h)} - e^{-yt}}{h}\right| \; = \; \left|-ye^{-y(t+\theta h)}\right| \; \le \; ye^{-\frac12yt} \hspace{2cm} y > 0$ (for some $0 < \theta < 1$ , while $\lim_{h \to 0}\frac{e^{-y(t+h)} - e^{-yt}}{h} \; = \; -ye^{-yt} \hspace{2cm} y > 0$ Thus the DCT implies that $\lim_{h \to 0} \frac{I(t+h) - I(t)}{h} \; = \; \lim_{h \to 0} \int_0^\infty \frac{e^{-y(t+h)} - e^{-yt}}{h} \frac{\sin y}{y}\,dy \; = \; -\int_ 0^\infty \sin y \,e^{-yt}\,dy$ which proves differentiation under the integral sign here. As you say, this still leaves the matter of the continuity of $I$ at $0$ .

To be honest, it is easier to evaluate the improper integral $\frac{\sin x}{x}$ by integrating $\frac{e^{iz}}{z}$ around the almost semicircular contour of radius $R$ in the upper half plane, with a small semicircular contour of radius $\varepsilon$ avoiding the singularity at $0$ . The fact that

$\int_0^\infty \frac{\sin x}{x}\,dx \; = \; \lim_{{R \to \infty} \atop {\varepsilon \to 0}} \int_\varepsilon^R \frac{\sin x}{x}\,dx \; = \; \tfrac12\pi$ is a matter of standard contour integration.

Continuity from above at $t=0$ follows from "Leibniz' Convergence Criterium". Let $\varepsilon>0$ and choose

• $n\in\mathbb{N}$ large enough such that $\frac{2}{n\pi}<\frac{\varepsilon}{3}$
• $t\geq 0$ small enough such that $1-e^{-n\pi t}<\frac{\varepsilon}{3\pi n}$

Then we use the "triangle inequality" to estimate $\begin{aligned} |I(0)-I(t)|&\leq \left| \int_{ n\pi}^{\infty}\frac{\sin(x)}{x}\:dx \right|+\left| \int_{n\pi}^\infty\frac{\sin(x)}{x}e^{-xt}\:dx \right|+\int_0^{n\pi}\left|\frac{\sin(x)}{x}\right|(1-e^{-xt})\:dx \end{aligned}$ With $|\frac{\sin(x)}{x}|\leq 1$ and the choice for $t$ the third integral is smaller than $\frac{\varepsilon}{3}$ . We rewrite the first two integrals as alternating sums. A substitution $y:=x-k\pi$ yields $\begin{aligned} \int_{n\pi}^\infty\frac{\sin(x)}{x}e^{-xt}\:dx &= \sum_{k=n}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{\sin(x)}{x}e^{-xt}\:dx = \sum_{k=n}^{\infty}(-1)^k\int_0^{\pi}\frac{\sin(y)}{y+k\pi}e^{-(y+k\pi)t}\:dy=:\sum_{k=n}^{\infty}(-1)^k a_k\\[.5em] 0&<a_k<\int_0^{\pi}\frac{\sin(y)}{k\pi}\cdot 1\:dy=\frac{2}{k\pi} \end{aligned}$ The sequence $a_k$ is monotonically decreasing and converges to zero. By "Leibniz' Convergence Criterium" for alternating series, we get $t\geq 0:\qquad \left|\int_{n\pi}^{\infty} \frac{\sin(x)}{x}e^{-xt}\:dx\right|=\left|\sum_{k=n}^\infty (-1)^ka_k\right|<|a_n|<\frac{2}{n\pi}<\frac{\varepsilon}{3}$ Putting everything together, we have $|I(0)-I(t)|<\varepsilon$

Like I said, rather than all this, just use contour integration.