Tricky Integral with parameter with difficult limits

Question

I had to calculate:

$$\int_0^\infty\textrm{d}x\,\frac{1-\exp(-t)}{t}\cos(t)$$

My approach was to define a function with parameter:

$$\mathcal{F}(a)=\int_0^\infty\textrm{d}x\,\frac{1-\exp(-a t)}{t}\cos(t):=\int_0^\infty\textrm{d}x\,f(x,a)$$

Then we can consider:

$$\partial_af(x,a)=\exp(-a t) \cos(t)$$

Which is clearly dominated for $a\in[a_0,\infty[$ (for $a_0>0$) by a function: $\exp(-a_0t)$. Then there is a theorem that states that if there exists some $a_1$ for which:

$$\mathcal{F}(a_1)\; \textrm{is well defined} \quad \textrm{and}\quad \partial_af(x,a)\;\textrm{is dominated}$$

Then $\mathcal{F}(a)$ can be redefined to:

$$\mathcal{F}(a)=\mathcal{F}(a_1)+\int_{a_1}^{a}\textrm{d}y\int_0^\infty\textrm{d}x\;\partial_yf(x,y)$$

Which is then well defined on $a\in[a_0,\infty[$.

Then by exact calculation:

$$\int_0^\infty\textrm{d}x\;\exp(-a t) \cos(t)=\frac{a}{1+a^2}$$

And now integrating I finally got expression:

$$\mathcal{F}(a)=\frac{1}{2}\log(1+a^2)+C$$

And here emerges the problem to find the value of $C$. Then the final integral would be for $\mathcal{F}(1)$.

With Mathematica I got that $\mathcal{F}(1)=\frac{1}{2}\log(2)$ therefore $C=0$ and if someone could prove that:

$$\lim_{a\to0}\int_0^\infty\textrm{d}x\;f(x,a)=\int_0^\infty\textrm{d}x\;\lim_{a\to0}f(x,a)$$

I would be very grateful, because then $\lim_{a\to0}\;f(x,a)=0$ and $C=0$.

Answer 1

For any $a_1>0$, we get from your formulas

$F(a) = F(a_1) + \frac{1}{2}\ln(1+a^2) - \frac{1}{2}\ln(1+a_1^2)$.

Now letting $a_1 \to 0$, using continuity of $F(a)$ and obvious fact that $F(0)=0$, we get the following.

$F(a) = \frac{1}{2}\ln(1+a^2)$

Hence, $F(1)=\frac{1}{2}\ln(2)$.

Answer 2

Let $F(a)=\int_0^{\infty} \frac{1-e^{-a t}}{t} \cos t d t, \textrm{ where } a>0$, then $$ \begin{aligned} F^{\prime}(a) & =\int_0^{\infty} e^{-a t} \cos t d t =\frac{a}{a^2+1} \\ F(1)&=F(1)-\lim_{t\rightarrow 0}F(t) =\int_0^1 \frac{a}{a^2+1} d a=\frac{1}{2} \ln 2 \\ \int_0^{\infty} \frac{1-e^{-t}}{t} \cos t d t & =\frac{1}{2} \ln 2 \end{aligned} $$

Alternative method

Noting that $\int_0^1 e^{-x t} d t=\frac{1-e^{-t}}{t}$, we can change it into a double integral $$ \begin{aligned} I & =\int_0^{\infty}\left(\int_0^1 e^{-x t} d x\right) \cos t d t \\ & =\int_0^1\left(\int_0^{\infty} e^{-x t} \cos t d t\right) d x \\ & =\int_0^1 \frac{x}{x^2+1} d x \quad \textrm{(via IBP) } \\ & =\frac{1}{2}\left[\ln \left(x^2+1\right)\right]_0^1 \\ & =\frac{1}{2} \ln 2 \end{aligned} $$