A propos of a user's comment on this question, quoting Feynman to the effect that some integrals are only possible using contour integration, I wonder what the simplest example of such an integral might be. In particular, he spoke of integrals that were the complex part of a solution, divorced from the rest of the solution.

For me, something like

$$I = \int_0^\infty \frac{\ln x}{x^2+1}dx, $$

seems hard; it falls out as the complex part of contour integration using the residue theorem in the problem,

$$J = \int_0^\infty \frac{(\ln x)^2}{x^2+1}dx . $$ That is, we can use contour integration to evaluate J and end up with something like

$J + iI = \frac{\pi^3}{8}$ so we conclude $I = 0.$

Perhaps it is hard but I don't know that it can't be done without complex analysis [edit: Robjohn has shown it can be done without complex analysis]. So I would like to see one good example of the sort of thing Feynman might have had in mind.

Here is the original quote: "So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!"

In terms of this quote, do we have an example of what Paul did?


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    $\begingroup$ $I$ is not the best example of what you want. It is easily seen to be $0$ after one splits the interval to [0,1] and [1,infinity] and applying x=1/u to either of them. $\endgroup$ – Ragib Zaman Jul 6 '12 at 3:42
  • $\begingroup$ Exactly. It is not, as Robjohn showed. Can you think of a good one? $\endgroup$ – daniel Jul 6 '12 at 11:53
  • $\begingroup$ I am having difficulty ascertaining precisely what you are asking. Does my first answer apply? Equating the real and imaginary parts of $(5)$ and $(6)$ give $(7)$ and $(9)$. $\endgroup$ – robjohn Jul 6 '12 at 12:25
  • $\begingroup$ @robjohn: I re-edited and inserted the original quote. I also added the way in which your first answer was responsive to the question. Please let me know if still not clear. Thanks. $\endgroup$ – daniel Jul 6 '12 at 12:52
  • $\begingroup$ I'm pretty skeptical of the literal truth of Feynman's claim. Both Feynman integration and contour integration can be thought of as clever -- and fairly general -- ways of using Green's theorem to turn hard integrals into easy integrals, so intuitively I would expect them to be capable of doing roughly the same things. I suspect that Paul's integral was simply very difficult to do by differentiating under the integral sign. $\endgroup$ – Micah Jul 6 '12 at 19:13

I believe that I misunderstood the question in my first answer. What I think you want is a way to compute $$ \int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x $$ without using complex analysis.

With the substitution $x\mapsto\frac1x$, we get $$ \int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x=-\int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x $$ which says that $$ \int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x=0 $$

  • $\begingroup$ I edited the question quite a bit in response to your comment. This answer is helpful and responsive. The other aspect of the question is described in the quote. Hopefully it's clearer now. $\endgroup$ – daniel Jul 6 '12 at 13:03

A change of variables yields $$ \begin{align} \int_0^\infty\frac{\log(x)^2}{1+x^2}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{t^2}{1+e^{-2t}}e^-t\,\mathrm{d}t\\ &=2\int_0^\infty\frac{t^2}{1+e^{-2t}}e^{-t}\,\mathrm{d}t\\ &=2\int_{-\infty}^\infty t^2\left(e^{-t}-e^{-3t}+e^{-5t}-e^{-7t}+\dots\right)\,\mathrm{d}t\\ &=4\left(1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots\right)\tag{1} \end{align} $$ Define $$ \xi(n)=\sum_{k=0}^\infty(-1)^k\frac{1}{(2k+1)^n}\tag{2} $$ and consider the generating function of $\xi(n)$ for odd $n$: $$ \begin{align} \sum_{n=0}^\infty\sum_{k=0}^\infty(-1)^k\frac{z^{2n+1}}{(2k+1)^{2n+1}} &=\sum_{k=0}^\infty(-1)^k\frac{\frac{z}{2k+1}}{1-\left(\raise{1pt}{\frac{z}{2k+1}}\right)^2}\\ &=\frac{z}{2}\sum_{k=0}^\infty(-1)^k\left(\frac{1}{z+2k+1}-\frac{1}{z-2k-1}\right)\\ &=\frac{z}{2}\sum_{k=-\infty}^\infty\left(\frac{1}{z+4k+1}-\frac{1}{z+4k-1}\right)\\ &=\frac{z}{8}\sum_{k=-\infty}^\infty\left(\frac{1}{k+\frac{z+1}{4}}-\frac{1}{k+\frac{z-1}{4}}\right)\\ &=\frac{z}{8}\left(\pi\cot\left(\pi\frac{z+1}{4}\right)-\pi\cot\left(\pi\frac{z-1}{4}\right)\right)\\ &=\frac{\pi z}{4}\sec\left(\frac{\pi z}{2}\right)\\ &=\frac{\pi z}{4}+\frac{\pi^3z^3}{32}+\frac{5\pi^5z^5}{1536}+\dots\tag{3} \end{align} $$ where we used $(7)$ from this answer.

Putting together $(1)$, $(2)$, and $(3)$ yields $$ \int_0^\infty\frac{\log(x)^2}{1+x^2}\,\mathrm{d}x=\frac{\pi^3}{8}\tag{4} $$

Here is another method using contour integration.

In order to define $\log(z)$, consider the path $\gamma$ that runs from $0$ to $\infty$ just north of the real axis, circles the complex plane counter-clockwise, then returns from $\infty$ to $0$ just south of the real axis.

Adding the residues at $z=i$ and $z=-i$ gives $$ \begin{align} \int_\gamma\frac{\log(z)^3}{1+z^2}\mathrm{d}z &=2\pi i\left(\frac{\left(\frac\pi2i\right)^3}{2i}+\frac{\left(\frac{3\pi}2i\right)^3}{-2i}\right)\\ &=\frac{13\pi^4}{4}i\tag{5} \end{align} $$ Computing the integral along the real axis yields $$ \begin{align} \int_\gamma\frac{\log(z)^3}{1+z^2}\mathrm{d}z &=\int_0^\infty\frac{\log(x)^3-(\log(x)+2\pi i)^3}{1+x^2}\mathrm{d}x\\ &=\int_0^\infty\frac{-3\log(x)^22\pi i+\color{#C00000}{3\log(x)4\pi^2}+\color{#00A000}{8\pi^3i}}{1+x^2}\mathrm{d}x\\ &=-6\pi i\int_0^\infty\frac{\log(x)^2}{1+x^2}\mathrm{d}x+\color{#C00000}{0}+\color{#00A000}{4\pi^4i}\tag{6} \end{align} $$ Equating the real and imaginary parts of $(5)$ and $(6)$ gives $$ \int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x=0\tag{7} $$ and $$ -6\pi i\int_0^\infty\frac{\log(x)^2}{1+x^2}\mathrm{d}x=-\frac{3\pi^4}{4}i\tag{8} $$ which yields $$ \int_0^\infty\frac{\log(x)^2}{1+x^2}\mathrm{d}x=\frac{\pi^3}{8}\tag{9} $$


Shortest possible answer would be Split the integral into intervals from $[0,1]$ and then $[1,\infty]$ and in the second integral make substitution $x=\frac{1}{t}$ and this all reduces to 0


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