Integration with a little bit of Leibnitz rule

Question

The problem is simple. But the solution is where I get my doubts. Also I would appreciate if anyone could fill me in with all the necessary points where I need to check for convergation for the procedure to correct.

$$\int_0^\infty \frac{\cos(bx)-\cos(cx)}x e^{-ax} \,dx $$ for all $a,b,c \in \rm I\!R ; a>0$

From here it's obvious that we can rewrite this as $$\frac d{dt} \left.\int_0^\infty \frac{cos(tx)}x e^{-ax}\right|_c^b \,dx =\int_0^\infty \int_c^b {\sin(tx)} e^{-ax} \,dx \,dt= \int_c^b I\,dt$$

Then via double per partes $$I=\int_0^\infty {\sin(tx)} e^{-ax}\,dx = \left.\frac{-1}{a}\sin(tx)e^{-ax}\right|_{t=0}^\infty + \frac ta \int_0^\infty {cos(tx)} e^{-ax}dx =$$ $$=\left. \frac{-\sin(tx)e^{-ax}}a \right|_{t=0}^\infty - \left. \frac t a {\cos(tx)e^{-ax}}\right|_{t=0}^\infty - \frac{t^2}{a^2}\int_0^\infty {\sin(tx)} e^{-ax}\,dx$$ $$I=\frac ta-\frac{t^2}{a^2}I$$ $$I=\frac{\frac ta}{1+\frac{t^2}{a^2}}$$ $$\int_c^b\frac{\frac ta}{1+\frac{t^2}{a^2}}=\int_{c/a}^{b/a}\frac{u}{1+u^2}=\frac12\ln(\frac dc)$$

I find it odd that my final result is not dependant at all of parameter $a$ Just does not sound right.

And I would appreciate filling out where covnergation or any other possible breaking points need to be checked.

EDIT: A typo...probably more to come.

Answer 1

In the OP, there was a flaw in the beginning of the development. The integral of interest is $I(a)=\int_0^\infty \frac{\cos(bx)-\cos(cx)}{x}e^{-ax}\,dx$.

But, $I(a)$ is not equal to $\left.\left(\frac{d}{dt}\int_0^\infty \frac{\cos(tx)}{x}e^{-ax}\,dx\right)\right|_{c}^{b}$. In fact, this latter integral fails to exist due to the $\frac1x$ singularity at $x=0$.

Herein, we present two methodologies that we can use to evaluate $I(a)$.

---

NOTE: Since the integral is even in $b$ and $c$, we may assume without loss of generality that $b\ge 0$ and $c \ge 0$.

---

METHODOLOGY $(1)$: Differentiating Under the Integral Sign

Here is a straightforward approach. Let $I(a)$ be given by

$$I(a)=\int_0^\infty \frac{\cos(bx)-\cos(cx)}{x}e^{-ax}\,dx$$

Differentiating reveals

$$I'(a)=\int_0^\infty e^{-ax}\left(\cos(cx)-\cos(bx)\right)\,dx=\frac{a}{a^2+c^2}-\frac{a}{a^2+b^2}$$

Integrating $I'(a)$, we obtain

$$I(a)-I(0)=\int_0^a \left(\frac{t}{t^2+c^2}-\frac{t}{t^2+b^2}\right)\,dt=\frac12\log\left(\frac{a^2+c^2}{a^2+b^2}\right)-\log(c/b)$$

Using a slightly modified version of Frullani's Theorem, we find that $I(0)=\log(c/b)$. Therefore,

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\cos(bx)-\cos(cx)}{x}e^{-ax}\,dx=\frac12\log\left(\frac{a^2+c^2}{a^2+b^2}\right)}$$

---

METHODOLOGY $(2)$: Applying Frullani's Theorem for Complex Parameters

In THIS ANSWER, I showed that if $f(z)$ is analytic, then the following generalization of Frullani's Theorem holds for complex parameters $a$ and $b$:

$$\begin{align} \int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx&=f(0)\log(|b/a|)\\\\ &+if(0)\left(\arctan\left(\frac{\text{Re}(a\bar b)-|a|^2}{\text{Im}(a\bar b)}\right)-\arctan\left(\frac{|b|^2-\text{Re}(a\bar b)}{\text{Im}(a\bar b)}\right)\right) \tag 1 \end{align}$$

Then, we can write

$$I(a)=\text{Re}\left(\int_0^\infty \frac{e^{-(a-ib)x}-e^{-(a-ic)x}}{x}\,dx\right)$$

Applying $(1)$ with $f(z)=e^{-z}$, $a\to a-ib$ and $b\to a-ic$ yields

$$\begin{align} I(a)&=e^{-(0)}\log\left(\frac{\sqrt{a^2+c^2}}{\sqrt{a^2+b^2}}\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac12\log\left(\frac{a^2+c^2}{a^2+b^2}\right)} \end{align}$$

as expected!

Answer 2

I get:

$\frac {d}{dt}\int_0^\infty \frac{\cos(tx)}x e^{-ax}|_c^b dx \\ \int_0^\infty \int_c^b {\sin(tx)} e^{-ax} dt dx\\ \int_c^b\int_0^\infty {\sin(tx)} e^{-ax} dxdt\\ \int_c^b \frac {t}{a^2+t^2} dt\\ \frac 12 (\ln(a^2 + b^2) - \ln(a^2 + c^2))$