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I am not able to prove that

$$ \int_0^\infty \frac {e^{-\alpha x}-e^{-\beta x}}{x} \sin(\gamma x)\, \mathrm{ d}x = \arctan\left( \frac {\beta}{\gamma} \right) - \arctan\left( \frac {\alpha}{\gamma} \right) \tag{**} $$

for $ \alpha, \beta, \gamma \in \mathbb R, \gamma \neq 0 $.

However, I managed to find a very clever proof for

$$ \int_0^\infty \frac {e^{-\alpha x}-e^{-\beta x}}{x}\, \mathrm{d}x = \ln\left( \frac { \beta }{ \alpha } \right) \tag{*} $$

using the improper integral $I(y) = \int_0^\infty e^{-yx}\, \mathrm{d}x$.

The idea is the following. Because $I(y)$ converges uniformly in respect to $y$ we can integrate $I(y)$ from $\alpha$ to $\beta$ in respect to $y$ and we know the following equality: $$ \int_\alpha^\beta \left(\int_0^\infty e^{-yx}\, \mathrm{d}x\right)\,\mathrm{d}y = \int_0^\infty \left(\int_\alpha^\beta e^{-yx}\, \mathrm{d}y\right)\,\mathrm{d}x $$

By simply integrating the left and right members of this equality we prove the $(*)$ equality.

Any tips on how can I use the same technique for $(**)$.
I cannot wrap my head around it because the integral depends on three parameters, not two.
Is there another way to solve it ?

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  • $\begingroup$ try $I(y,\gamma) = \int_0^{\infty} e^{-yx} sin(\gamma x)d x$, you will get something like $\frac{\gamma}{y^2+\gamma^2}$ $\endgroup$ – Yimin Feb 15 '13 at 21:45
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    $\begingroup$ Thanks ! I managed to do it. $\endgroup$ – user54549 Feb 15 '13 at 22:04
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Note that: $$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

Therefore:

$$\begin{aligned}\int_0^\infty \frac{e^{-\alpha x}-e^{-\beta x}}{x}\sin(\gamma x) \ dx &= \int_0^\infty \left( \frac{e^{-\alpha x}-e^{-\beta x}}{x}\right)\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(\gamma x)^{2n+1} \ dx \\&= \sum_{n=0}^\infty \frac{(-1)^{n}\gamma^{2n+1}}{(2n+1)!} \left\{ \int_0^\infty e^{-\alpha x}x^{2n}dx-\int_0^\infty e^{-\beta x}x^{2n}dx\right\} \\ &= \sum_{n=0}^\infty\frac{(-1)^{n}\gamma^{2n+1}}{(2n+1)!} \left\{ \frac{\Gamma(2n+1)}{\alpha^{2n+1}}-\frac{\Gamma(2n+1)}{\beta^{2n+1}}\right\} \\&=\sum_{n=0}^\infty\frac{(-1)^{n}\gamma^{2n+1}}{(2n+1)!} \left\{ \frac{(2n)!}{\alpha^{2n+1}}-\frac{(2n)!}{\beta^{2n+1}}\right\} \\&=\sum_{n=0}^\infty \frac{(-1)^n \left( \frac{\gamma}{\alpha}\right)^{2n+1}}{2n+1} - \sum_{n=0}^\infty \frac{(-1)^n \left( \frac{\gamma}{\beta}\right)^{2n+1}}{2n+1} \\&=\arctan \left( \frac{\gamma}{\alpha}\right)-\arctan \left( \frac{\gamma}{\beta}\right)\end{aligned}$$

In the last step, I have used

$$\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$

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Here is an approach, $$I(\gamma)=\int_0^\infty \frac {e^{-\alpha x}-e^{-\beta x}}{x} \sin(\gamma x) dx\implies \frac{dI(\gamma)}{dx}=\int_0^\infty ({e^{-\alpha x}-e^{-\beta x}}) \cos(\gamma x) dx.$$

Now, the last integral is the Laplace transforms of $\cos(x)$ with respect to $\alpha$ and $\beta$. Can you finish the problem now?

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  • $\begingroup$ Thank you ! I already did after what @Yimin told me. I did not think about considering $ \gamma $ a constant. I guess I was too fixed on one idea. $\endgroup$ – user54549 Feb 15 '13 at 22:02
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If you put $I(y,\gamma) = \int_0^{\infty} e^{-yx}sin(\gamma x)\mathrm{d}x$, then you problem is equivalent to compute $$\int_{\alpha}^{\beta} I(y,\gamma)\mathrm{d}y$$ by using Fuini Theorem.

And $I(y,\gamma)=\frac{\gamma}{\gamma^2+y^2}$, thus $\int_{\alpha}^{\beta}\frac{\gamma}{\gamma^2+y^2} = \arctan(\gamma/y)|_{\alpha}^{\beta}$.

Done.

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I would just like to say that integral $$\int_0^\infty \frac {e^{-\alpha x}-e^{-\beta x}}{x}\, \mathrm{d}x = \ln\left( \frac { \beta }{ \alpha } \right) $$ which OP mentioned is Frullani's integrals:

Let $f:(0,+\infty) \to \mathbf{R}$ be locally integrable function, such that exists $\displaystyle f(0+)=\lim_{x \to 0+}f(x)$ and $\displaystyle f(+\infty)=\lim_{x \to +\infty}f(x)$. Then for $a,b >0$ it is true equality $$\int_0^{+\infty} \frac{f(bx)-f(ax)}{x}dx=(f(+\infty)-f(0+))\ln\left(\frac{b}{a}\right)$$

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