## Monday, January 1, 2007

### 2K7. Topic: Calculus. Level: Olympiad.

Problem: (2005 Putnam - A5) Evaluate $\displaystyle \int_0^1 \frac{\ln{(x+1)}}{x^2+1} dx$.

Solution: This seemingly harmless integral will turn into a beast in a moment, but first, the Leibniz Integral Rule. This states that, for a multivariate function $f(x, t)$, we have

$\displaystyle \frac{\partial}{\partial t} \int_{a(t)}^{b(t)} f(x, t) dx = \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t} dx + f(b(t), t) \frac{\partial b}{\partial t} - f(a(t), t) \frac{\partial a}{\partial t}$.

The neat thing is, when the bounds $a(t)$ and $b(t)$ are constant, the last two terms go away! So we basically just take the derivative of the function inside the integral. Anyway, so how does this apply to our problem? Well, our function seems to be univariate at the moment, but since we failed to solve it in its current form, let's complicate things. Suppose

$f(x, t) = \frac{\ln{(tx+1)}}{x^2+1}$ and let $\displaystyle I(t) = \int_0^1 f(x, t) dx$.

We want to evaluate $I(1)$. Applying the Leibniz Integral Rule, we obtain

$\displaystyle I^{\prime}(t) = \frac{\partial}{\partial t} \int_0^1 f(x, t) dx = \int_0^1 \frac{\partial f}{\partial t} dx$

$\displaystyle I^{\prime}(t) = \int_0^1 \frac{x}{(tx+1)(x^2+1)} dx$.

Using partial fractions, this becomes

$\displaystyle I^{\prime}(t) = \int_0^1 \left(-\frac{t}{t^2+1} \cdot \frac{1}{tx+1}+\frac{1}{t^2+1} \cdot \frac{x}{x^2+1}+\frac{t}{t^2+1} \cdot \frac{1}{x^2+1} \right) dx = \left[ -\frac{\ln{(tx+1)}}{t^2+1}+\frac{\ln{(x^2+1)}}{2(t^2+1)}+\frac{t}{t^2+1} \cdot \arctan{x} \right]_0^1$

$\displaystyle I^{\prime}(t) = -\frac{\ln{(t+1)}}{t^2+1}+\frac{\ln{2}}{2(t^2+1)}+\frac{\pi}{4} \cdot \frac{t}{t^2+1}$.

So it remains to recover $I$ through solving the differential equation

$\displaystyle I^{\prime}(t) = -\frac{\ln{(t+1)}}{t^2+1}+\frac{\ln{2}}{2(t^2+1)}+\frac{\pi}{4} \cdot \frac{t}{t^2+1}$.

Integrating both sides, we get

$\displaystyle I(t) = \int_0^t \left(-\frac{\ln{(t+1)}}{t^2+1}+\frac{\ln{2}}{2(t^2+1)}+\frac{\pi}{4} \cdot \frac{t}{t^2+1}\right) dt = -\int_0^t \frac{\ln{(t+1)}}{t^2+1} dt + \frac{\ln{2}}{2}\arctan{t} + \frac{\pi}{8} \ln{(t^2+1)}$

$\displaystyle I(1) = -\int_0^1 \frac{\ln{(t+1)}}{t^2+1} dt + \frac{\pi}{4} \ln{2}$.

But wait, recall that $\displaystyle I(1) = \int_0^1 \frac{\ln{(x+1)}}{x^2+1} dx$, so in fact the previous equation yields

$\displaystyle I(1) = -I(1) + \frac{\pi}{4} \ln{2} \Rightarrow I(1) = \frac{\pi}{8} \ln{2}$. QED.

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Comment: This probably wasn't the cleanest solution to this integral. Other solutions involved the substitutions $x = \tan{\theta}$, $x = \frac{1-u}{1+u}$, as well as some crazy series expansion. This technique seems to be well-known though, associated with the famous physicist Richard Feynman as one of his favorite tricks. It can be very helpful in evaluating otherwise difficult to approach integrals.

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Practice Problem: Evaluate $\displaystyle \int_0^1 \frac{x-1}{\ln{x}} dx$.