## Another Integral July 25, 2009

Posted by Dennis in Problem-solving.
Tags: ,

This is a problem from the Putnam competition I saw two years ago as a freshman, and I did it during my stats class this year. It uses symmetry in a similar way to Akhil’s post.
Find ${\displaystyle\int_{0}^{1}{\frac{\ln(x+1)}{x^2+1}dx}}$.
Let ${\tan(u)=x}$, then

$\displaystyle \begin{array}{rcl} \frac{du}{dx}=\frac{1}{1+x^2}. \end{array}$

and

$\displaystyle \begin{array}{rcl} \int_{0}^{1}{\frac{\ln(x+1)}{x^2+1}dx}=\int_{0}^{\frac{\pi}{4}}{\ln(\tan(u)+1)du}. \end{array}$

Also, we have the identity

$\displaystyle \begin{array}{rcl} \sin(u)+\cos(u)&=&\sqrt{2}(\sin(\frac{\pi}{4})\cos(u)+\cos(\frac{\pi}{4})\sin(u))\\&=&\sqrt{2}(\sin(u+\frac{\pi}{4})) \end{array}$

So

$\displaystyle \begin{array}{rcl} \int_{0}^{\frac{\pi}{4}}{\ln(\tan(u)+1)du}&=&\int_{0}^{\frac{\pi}{4}}{\ln(\frac{\sin(u)+\cos(u)}{\cos(u)})du}\\ &=&\int_{0}^{\frac{\pi}{4}}{\ln(\sin(u)+\cos(u))-\ln(\cos(u))du}\\ &=&\int_{0}^{\frac{\pi}{4}}{\ln(\sqrt{2}(\sin(u+\frac{\pi}{4})))-\ln(\cos(u))du}\\ &=&\int_{0}^{\frac{\pi}{4}}{\frac{\ln(2)}{2}+\ln(\cos(\frac{\pi}{4}-u))-\ln(\cos(u))du}\\ &=&\frac{\pi}{4}\frac{\ln(2)}{2}\\ &=&\frac{\pi\ln(2)}{8} \end{array}$

1. Qiaochu Yuan - July 25, 2009

The differentiating-under-the-integral-sign solution here is also quite nice: let $I(t) = \int_{0}^{1} \frac{\ln (tx + 1)}{x^2 + 1} \, dx$. Then

$\displaystyle I'(t) = \int_{0}^{1} \frac{x}{(tx + 1)(x^2 + 1)} \, dx$

and after a routine computation with partial fractions one finds

$\displaystyle I'(t) = \frac{\pi t}{4(t^2 + 1)} + \frac{\log 2}{2(t^2 + 1)} - \frac{\log (t + 1)}{t^2 + 1}$

hence, rather miraculously after integrating and simplifying,

$\displaystyle I(1) = \frac{\pi \log 2}{4} - I(1).$

The tangent substitution is the first official solution. The second official solution uses the substitution $x = \frac{1 - u}{1 + u}$ instead, which is closely related. The solution I just described is the third official solution, and the fourth is a fairly tedious but non-“magical” solution using series expansion.

2. Akhil Mathew - July 25, 2009

Two very clever solutions.

The Wikipedia page on differentiation under the integral sign has a few more examples. As far as I know, however, it isn’t a “standard” trick as normally one uses a contour integral.

3. liuxiaochuan - July 26, 2009

a small typo here:
it’s ‘$x=\tan u$‘，instead of $u\\tan x$

4. 两个快速积分 « Liu Xiaochuan’s Weblog - July 26, 2009

[…] 其二是 .首先令,则 […]

5. Dennis - July 26, 2009

liuxiaochuan,

Thanks for catching that, I fixed it.

6. Todd Trimble - July 27, 2009

How “standard” differentiation under the integral is depends on who you ask. As Qiaochu has said, Doron Zeilberger has made something of a theory of it (reference); it is connected with the method of “creative telescoping” which is used in Wilf-Zeilberger algorithms for evaluating hypergeometric sums.

I guess you could say that this method is not standard or fashionable in textbooks currently. I think it should be more widely known.

7. Akhil Mathew - July 27, 2009

Yes, that’s more or less what I meant to say; I am not aware of its being widely taught (though I am basing this claim on independent reading of textbooks, which I assume reflect the contents in the corresponding courses), even though it’s useful, as we discussed.