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I want to calculate:$$\int_0^{\infty}\frac{{e^{-ax}-e^{-bx}}}{x}\cos(cx)\mathrm{d}x\qquad(a,b,c > 0)\tag{1}$$And I want to use:$$\frac{x}{x^2+k^2}=\int_{0}^{\infty}e^{-xy}\cos(ky)\mathrm{d}y\tag{2}$$So, I want to think:$$\int_0^{\infty}\frac{{e^{-tx}}{\cos(cx)}}{x}\mathrm{d}x=\int_0^{\infty}\mathrm{d}x\int_t^{\infty}e^{-xy}\cos(cx)dy\tag{3}$$How can I use Fubini to chage the order of integration?

Math Attack
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Flat leaf
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2 Answers2

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\begin{align} &\int_0^{\infty}\frac{{(e^{-ax}-e^{-bx})}{\cos(cx)}}{x}dx\\ =& \int_0^{\infty}\int_a^b e^{-xy}\cos(cx)\ dy dx = \int_a^b \frac y{c^2+y^2}dy=\frac12\ln\frac{c^2+b^2}{c^2+a^2} \end{align}

Quanto
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  • How to say $\int_0^{\infty}\frac{{e^{-tx}}{\cos(cx)}}{x}dx=\int_0^{\infty}dx\int_t^{\infty}e^{-xy}\cos{cx}dy$is integrable.(how to show (3) < ∞？) – Flat leaf Jun 14 '22 at 02:53
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Using the Frullani' theorem and denoting $f(x)=e^{-x}$ $$\int_0^{\infty}\frac{(e^{-ax}-e^{-bx})\,\cos(cx)}{x}dx=\Re\int_0^{\infty}\frac{(e^{-ax-icx}-e^{-bx-icx})}{x}dx$$ $$=\Re\,\big(f(\infty)-f(0)\big)\ln\frac{a+ic}{b+ic}=\frac{1}{2}\ln\frac{b^2+c^2}{a^2+c^2}$$

J.G.
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Svyatoslav
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    Does Frullani work for complex numbers like that? – J.G. Jun 14 '22 at 05:53
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    @J.G. thank you for the good question. The generalisation of the Frullani' integral can be found here https://math.stackexchange.com/questions/1807410/frullani-s-theorem-in-a-complex-context. Also, for our specific case we can split the integral into two parts $\displaystyle \int_0^{\infty}\frac{(e^{-ax-icx}-e^{-bx-icx})}{x}dx=\int_0^{\infty}\frac{(e^{-ax-icx}-e^{-x})}{x}dx-\int_0^{\infty}\frac{(e^{-bx-icx}-e^{-x})}{x}dx$ and follow the way proposed by Jack D'Aurizio (second post). – Svyatoslav Jun 14 '22 at 07:21