Previously, I read The Integral that Stumped Feynman?. It was here that I found that there were certain complex valued integrals that even stumped Feynman's DUI. I used to believe that Differentiation Under the Integral was a powerful substitute for Contour Integration. In general, I am led to believe Contour Integration is the single most powerful technique of integration; however, I was wondering are there any integrals for which Contour Integration would fail and Differentiation Under the Integral would work in?

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1 Answers1


In my opinion, contour integration/residue theorem and DUI’s strengths lie in different areas.

For integrals of good limits, residue theorem can tackle problems that might be even impossible to solve by DUI.

e.g. $$\text{P}\int^\infty_{-\infty}\frac{\sin x}{\prod^n_{k=0}(x-a_k)}dx$$ for an arbitrary sequence of complex numbers $\{a_k\}$.

Also, residue theorem has a higher versatility. For instance, we can solve the Basel’s problem by residue theorem but not by DUI.

By considering $$\oint_{|z|=R}\frac{\pi\cot \pi z}{z^2}dz$$ and let $R\to\infty$, you can obtain $$\sum^\infty_{k=0}\frac1{k^2}=-\frac12\text{Res}_{z=0}\frac{\pi\cot \pi z}{z^2}$$

Using a similar method you can even derive a general formula for the value of zeta function at positive even numbers.

Plus, residue theorem can be applied in many more cases, like finding the inverse of Laplace transform and Mellin transform, or even derive some quite crazy formula like I did in one of my answers which I haven’t elaborated.

On the other hand, DUI is extraordinarily useful in tackling some integrals. Wikipedia has a bunch of them(just search Leibniz’s integral rule).

For example, I can’t see how can residue theorem directly tackle the integral $$\int^1_0\frac{a^2}{x^2+a^2}dx$$

Another good example is that DUI can solve, without any substitution, the integral $$\int^1_0\frac{x-1}{\ln x}dx$$ by considering the integral $$\int^1_0\frac{x^a-1}{\ln x}dx$$ Residue theorem can’t solve that directly.

Although I wrote a lot more for residue theorem, as I said at the very beginning, they are in general equally powerful when evaluating integrals. Both techniques are must-learns for a person who loves math.

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