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$$\int\limits_{0}^{1}\cos \left (\ln \frac{1}{x} \right )\frac{x^b-x^a}{\ln x}dx=\int\limits_{0}^{1}\frac{\cos \ln x}{\ln x}\left ( x^b-x^a \right )dx$$ $$t=-\ln x\Rightarrow dx=-e^{-t}dt\Rightarrow -e^t=x$$ $$\int\limits_{-\infty }^{0}\frac{\cos (-t)}{-t}\left ( e^{-tb}-e^{-ta} \right )e^tdt=-\int\limits_{-\infty }^{0}\frac{1}{t}\cdot \frac{e^{it}+e^{-it}}{2}\cdot \left ( e^{t-bt}-e^{t-at} \right )dt$$ How do I calculate here using the Frullani formula?

Saurav Maheshkar
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Dmitry
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    I would recommend the substitution $x=e^{-t}$. Then $$I=\int_0^\infty\frac{\cos t}{t}\big(e^{-at}-e^{-bt}\big)e^{-t}dt=\Re\int_0^\infty\frac{e^{-(a+i+1)t}-e^{-(b+i+1)t}}{t}dt$$ The Frullani integral method can be applied, but there is some specifics for complex numbers - https://math.stackexchange.com/questions/1807410/frullani-s-theorem-in-a-complex-context – Svyatoslav Jun 17 '22 at 06:52
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    Thank you! This post helped me a lot! – Dmitry Jun 17 '22 at 07:46
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    See this: https://math.stackexchange.com/questions/4472101/calculating-the-integeral-int-0-infty-frace-ax-e-bx-coscxx/ – River Li Jun 18 '22 at 03:06

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