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I'm very much a beginner at using this integration technique and this is my first attempt at tackling an integral blindly and choosing this method so I'm not even sure if it can be solved this way. But my attempt came sort of close to the actual answer so I'm interested to see what went wrong and if my choice of the variable $b$ can even lead to a correct answer. This is problem $555$ from Putnam and Beyond (in the text a different approach was used).

$$I=\int_{0}^{\infty} \frac{\ln x}{x^2 + a^2}\mathrm{d}x$$

Let $$I(b)=\int_{0}^{\infty} \frac{\ln (bx)}{x^2 + a^2}\mathrm{d}x$$ Note: I wasn't that big a fan of the placement of this variable so I'd be intertesed to see other placements.

\begin{align} I'(b)&=\int_{0}^{\infty} \frac{\partial}{\partial b}\frac{\ln (bx)}{x^2 + a^2}\mathrm{d}x \\ I'(b)&=\frac{1}{b}\int_{0}^{\infty} \frac{1}{x^2+a^2}\mathrm{d}x \\ I'(b)&=\frac{1}{ab}\left[\arctan\left(\frac{x}{a}\right)\right]^{\infty}_{0} \\ I'(b)&=\begin{cases}\frac{1}{ab}\left(\frac{\pi}{2}-0\right) &,a>0 \\\frac{1}{ab}\left(-\frac{\pi}{2}-0\right) &,a<0 \end{cases} \\ I(b)&=\begin{cases}\frac{\pi}{2a}\ln b+C_1 &,a>0 \\ -\frac{\pi}{2a}\ln b+C_2 &,a<0 \end{cases} \end{align} Also $\displaystyle I(1/x)=\int_{0}^{\infty} \frac{\ln (1)}{x^2+a^2}\mathrm{d}x=0 $, so $\displaystyle I(1/x)= \begin{cases}\frac{\pi}{2a}(-\ln x)+C_1 &,a>0 \\ \frac{\pi}{2a}(\ln x)+C_2 &,a<0\end{cases} \Leftrightarrow \begin{cases}C_1=\frac{\pi}{2a}\ln x &,a>0 \\ C_2=-\frac{\pi}{2a}\ln x &,a<0 \end{cases}$ and for $x=1\Rightarrow C_{1,2}=0$

$$I(b)=\pm\frac{\pi}{2a}\ln b$$

But $$I=I(1)=0$$ Which is wrong. But the correct answer felt familiar when I saw it; $\displaystyle I=\frac{\pi \ln a}{2a}$ so what went wrong here?

ericforman
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  • N.b. This integral can also be handled slickly with the substitution $x = \frac{a^2}{u}$. – Travis Willse May 28 '23 at 20:33
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    "What went wrong here" is $I(1/x)=\int_{0}^{\infty} \frac{\ln (y/x)}{y^2+a^2}\mathrm{d}y\ne\int_{0}^{\infty} \frac{\ln (1)}{y^2+a^2}\mathrm{d}y.$ – Anne Bauval May 28 '23 at 20:33
  • @Travis $x=\frac{a}{t}$ is also used in the solutions from the book. – ericforman May 28 '23 at 20:37
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    Wlog $a>0.$____ – Anne Bauval May 28 '23 at 20:37
  • Hint: if you know how to prove $\int_0^\infty\frac{x^{s-1}}{x^2+a^2}\mathrm{d}x=\frac{\pi}{2}a^{s-2}\csc\frac{\pi s}{2}$, show its $s$-derivative at $s=1$ is $\frac{\pi}{2a}\ln a$. – J.G. May 28 '23 at 20:38
  • @AnneBauval I see, I never made a substitution using a variable in that context before and while it felt sketchy, it made sense to me for some reason. – ericforman May 28 '23 at 20:42
  • Various "duplicates": $\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral – Anne Bauval May 28 '23 at 20:42
  • @J.G can this be done in elementary methods or differentiation under the integral considering $I(s)$? I'm not sure how to do this otherwise because I'm not familiar with intergal transforms and since this is a Putnam prep book, I'm not sure how to incorporate transforms in a Putnam proof. – ericforman May 28 '23 at 21:11
  • The starting point I suggested is equivalent to the Gamma function's reflection formula. Do you know how to prove that - by the residue theorem (which can also avoid differentiation under the integral sign) or otherwise - or take it as read? – J.G. May 28 '23 at 21:16
  • I can't find exact information on this but I think I could probably use the Gamma function without any more justification and then calculating $\Gamma(z)\Gamma(1-z)$. Thank you. – ericforman May 28 '23 at 22:04

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