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Can anyone tell me if I made any errors in my solution and whether or not my answer is correct? Is there a better/quicker method to solving this?

If you can't see the writing very well (and therefore can't see my work and answer very well), then note the problem is read as "the integral from 0 to infinity of $\cos(bx)(x-\log(e^x-1)~dx$ where $b>0$"

\begin{align} \require{cancel} \int_0^\infty e^{-ax}\cos(bx)~dx&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}\int_0^\infty\cos(bx)(bx)^{2n}~dx\\&=\sum_{n=0}^\infty\frac{(-b^2)^n}{(2n)!}\int_0^\infty e^{-ax}x^{2n}~dx \quad\boxed{u=ax,~du=a~dx,~x=\frac ua} \\ &= \frac1a\sum_{n=0}^\infty\frac{(-b^2)^n}{(2n)!(a^2)^n}\int_0^\infty u^{2n}e^{-u}~du\\&= \frac1a\sum_{n=0}^\infty\left(\frac{-b^2}{a^2}\right)^n\frac{\cancel{(2n)!}}{\cancel{(2n)!}}\\&= \frac1a\sum_{n=0}^\infty\left(\frac{-b^2}{a^2}\right)^n=\frac1a\left[\frac1{1+\frac{b^2}{a^2}}\right]=\frac a{a^2+b^2} \end{align}

\begin{align} \require{cancel} \int_0^\infty e^{-ax}\cos(bx)~dx=\frac a{a^2+b^2}\implies \frac1{\cancel a}\left[\frac{\cancel a}{a^2+b^2}\right]&=\frac1a\int_0^\infty e^{-ax}\cos(bx)~dx\\&= \sum_{a=1}^\infty\frac1{a^2+b^2}=\sum_{a=1}^\infty\int_0^\infty\frac{e^{-ax}}a~dx \end{align}

$$\implies \frac{\pi b\coth(\pi b)-1}{2b^2}=\int_0^\infty\cos(bx)\sum_{a=1}^\infty\frac{e^{-ax}}a~dx\implies \sum_{a=1}^\infty\frac{e^{-ax}}a=\frac{e^{-x}}{1-e^{-x}}=\frac1{e^x-1}$$

\begin{align} \implies-\sum_{a=1}^\infty\frac{e^{-ax}}a=\int\frac1{e^x-1}~dx&\implies\sum_{a=1}^\infty\frac{e^{-ax}}a=-(\ln(e^x-1)-x)=x-\ln(e^x-1)\end{align}

$$\implies\small\sum_{a=1}^\infty\frac{e^{-ax}}a=\int_0^\infty\cos(bx)\sum_{a=1}^\infty\frac{e^{-ax}}a~dx=\int_0^\infty\cos(bx)(x-\ln(e^x-1))~dx=\frac{\pi b\coth(\pi b)-1}{2b^2}$$

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    I apologize for any poorly written notation and/or poor etiquette. This is my first time on any website like this. So far pointguardo edited a few notational errors, than you! –  Aug 05 '18 at 12:16
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    I have edited the post to now use MathJax. Feel free to [edit] it if you see anything wrong, and please use MathJax in the future to the best of your ability. For lengthier posts, please use the sandbox. – Simply Beautiful Art Aug 05 '18 at 12:39
  • @SimplyBeautifulArt Didn't know about $\cancel{cancel}$ til today! Cool! – BCLC Aug 05 '18 at 13:08
  • It seems that you've use the symbol $\implies$ to indicate your route, i.e. not necessarily implications. If you could organize your expressions, maybe it would be more digestible for us…? Also rigorously speaking the infinite sum and the integral signs are not interchangeable in general, but in your solution they seems to be reasonable. – xbh Aug 05 '18 at 13:22
  • Yes, @xbh is correct. The conditional symbol is definitely not meant to indicate implication. On paper I have some personal habits of guiding myself in my own way. –  Aug 05 '18 at 13:28

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You want to double-check your correct conclusion that if $b>0$ then $$\int_0^\infty\cos bx \ln(1-e^{-x})dx=\frac{1-\pi b \coth \pi b}{2b^2}.$$Here's a slicker way: $$-\sum_{n\ge 1}\frac{1}{n}\Re \int_0^\infty\exp-(n-ib)xdx=-\sum_{n\ge 1}\frac{1}{n^2+b^2}.$$Finish with $$\coth \pi b=\frac{1}{\pi b}+\frac{2b}{\pi}\sum_n\frac{1}{n^2+b^2}.$$

J.G.
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