How can I evaluate $ \int_0^\infty \frac{\ln(2e^x-1)}{e^x-1} dx $ using Feynman's trick?

Asked Jun 10 '24 at 21:57

Active Jun 10 '24 at 21:59

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I would like to find out a solution using just Feynman's method and Calculus I basic methods, to share with a novice student. I have found at Quora and this site some solutions containing transcendental functions and infinite series, which may not help so much in this case.

edited Jun 10 '24 at 21:59

asked Jun 10 '24 at 21:57

Arthur

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Do you have reason t9 believe this integral has a closed form, or that Feynman's method can be used on this integral? – Thomas Andrews Jun 10 '24 at 22:06
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And if you've found the value on this site, link to one, and tell us what you know. – Thomas Andrews Jun 10 '24 at 22:07
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According to Wolfram, it has a closed form – Sine of the Time Jun 10 '24 at 22:11
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This question here has an answer that uses the Leibniz integral rule for differentiating under the integral sign. – Accelerator Jun 10 '24 at 22:15
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The YouTube channel Maths 505 did a great video on this exact question, you can find his solution here. https://www.youtube.com/watch?v=ayls_5VjcXQ – Aidan Hyde Jun 10 '24 at 22:25

How can I evaluate $ \int_0^\infty \frac{\ln(2e^x-1)}{e^x-1} dx $ using Feynman's trick?

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