# Solution Verification: Compute $\int_{0}^{+\infty} \frac{\mathrm{e}^{-a x}-\mathrm{e}^{-b x}}{x} \sin x \mathrm{~d} x$, where $b>a>0$

Compute $$\int_{0}^{+\infty} \frac{\mathrm{e}^{-a x}-\mathrm{e}^{-b x}}{x} \sin x \mathrm{~d} x,$$ where $$b>a>0$$

$$\forall y \in[a, b]$$, $$\left|\mathrm{e}^{-x y}\right| \leq \mathrm{e}^{-a x}$$, and since $$\int_{0}^{+\infty} \mathrm{e}^{-a x} \mathrm{~d} x$$ convergence, we have $$\int_{0}^{+\infty} \mathrm{e}^{-x y} \mathrm{~d} x$$ uniformly convergence with respect to $$y \in[a, b]$$.

Note that $$\frac{\mathrm{e}^{-a x}-\mathrm{e}^{-b x}}{x}=\int_{a}^{b} \mathrm{e}^{-x y} \mathrm{~d} y,$$ so $$\int_{0}^{+\infty} \frac{\mathrm{e}^{-a x}-\mathrm{e}^{-b x}}{x} \sin x \mathrm{~d} x=\int_{a}^{b} \mathrm{~d} y \int_{0}^{+\infty} \mathrm{e}^{-x y} \sin x \mathrm{~d} x,$$ where \begin{aligned} & \int_{0}^{+\infty} \mathrm{e}^{-x y} \sin x \mathrm{~d} x \\ = & -\frac{1}{y} \int_{0}^{+\infty} \sin x \mathrm{de}^{-x y} \\ = & -\frac{1}{y}\left(\left.\mathrm{e}^{-x y} \sin x\right|_{0} ^{+\infty}-\int_{0}^{+\infty} \mathrm{e}^{-x y} \cos x \mathrm{~d} x\right) \\ = & -\frac{1}{y}\left(0+\frac{1}{y} \int_{0}^{+\infty} \cos x \mathrm{de}^{-x y}\right) \\ = & -\frac{1}{y^{2}}\left(\left.\mathrm{e}^{-x y} \cos x\right|_{0} ^{+\infty}+\int_{0}^{+\infty} \mathrm{e}^{-x y} \sin x \mathrm{~d} x\right) \\ = & -\frac{1}{y^{2}}\left(-1+\int_{0}^{+\infty} \mathrm{e}^{-x y} \sin x \mathrm{~d} x\right). \end{aligned} Therefore, $$\int_{0}^{+\infty} \mathrm{e}^{-x y} \sin x \mathrm{~d} x=\frac{1}{1+y^{2}}.$$

Then

\begin{aligned} & \int_{0}^{+\infty} \frac{\mathrm{e}^{-a x}-\mathrm{e}^{-b x}}{x} \sin x \mathrm{~d} x \\ = & \int_{a}^{b} \mathrm{~d} y \int_{0}^{+\infty} \mathrm{e}^{-x y} \sin x \mathrm{~d} x \\ = & \int_{a}^{b} \frac{1}{1+y^{2}} \mathrm{~d} y \\ = & \arctan b-\arctan a . \end{aligned}

Is my solution correct? Did I make careless mistakes in my solution? Thank you very much!

• Looks good to me! Mar 5 at 22:20

Below is an alternative approach, leading to the same result. $$I(t)=\int_0^\infty \frac{e^{-a x}-e^{-b x}}{x} \sin (t x) \ dx$$ $$I’(t)= \int_0^\infty ({e^{-a x}-e^{-b x}}{x} )\cos (t x) \ dx =\frac a{a^2+t^2} - \frac b{b^2+t^2}$$ Then $$\int_0^\infty \frac{e^{-a x}-e^{-b x}}{x} \sin x \ dx =\int_0^1 I’(t)dt\\ = \int_0^1 \frac a{a^2+t^2} - \frac b{b^2+t^2} \ dt = \tan^{-1}b-\tan^{-1}a$$
Let a and b are real numbers $$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\sin x\space dx=\Im\int_0^\infty\frac{e^{-(a-i)x}-e^{-(b-i)x}}{x}dx$$ Using the Frulliani integral formula we get $$I=\Im \ln\left(\frac{b-i}{a-i}\right)=\Im \ln\left(\frac{a+i}{b+i}\right)=\arctan(1/a)-\arctan(1/b)=\arctan b-\arctan a$$ We get the same answer