# Help with integrating 101: $\int y \ln{y}\,\mathbb dy$

Would appreciate it if someone would please help me solve this

$$\int y\;\ln y\, \mathbb dy$$

taking time to explain reason for each step taken.

Thanks in advance!

• Proceed by integration by parts Click show steps :) – mathmath8128 Jul 10 '11 at 1:45
• Why is there a dot in the integral? – user9464 Jul 10 '11 at 2:00
• @Jack: some people write multiplication of $a$ with $b$ as a period: $a.\!b$ (it is very common in France). – t.b. Jul 10 '11 at 2:05
• @Theo: +1. I've never learned this before. Nice to know that. – user9464 Jul 10 '11 at 2:23
• This technique is also useful for integrating with inverse trigonometric functions. – ncmathsadist Jul 10 '11 at 2:30

## 4 Answers

Use integration by parts:

$$\int \frac{df}{dy}\ g \ \ \text{d}y = fg - \int f \ \frac{dg}{dy} \ \ \text{d}y$$

Thus we get

$$\int \frac{d (y^2/2)}{dy}\ \ln y \ \ \text{d}y = \frac{y^2}{2}\ \ln y - \int \frac{y^2}{2} \frac{d \ln y}{dy} \ \ \text{d}y$$

Now $\displaystyle \frac{d \ln y}{dy} = \frac{1}{y}$.

The last integral becomes

$$\int \frac{y^2}{2} \frac{1}{y} \ \ \text{d}y = \int \frac{y}{2} \ \ \text{d}y = \frac{y^2}{4}$$

Thus your integral is

$$\frac{y^2 \ln y}{2} - \frac{y^2}{4} + C$$

The reason to use integration by parts is that the derivative of $\displaystyle \ln y$ is $\displaystyle \frac{1}{y}$ which cancels out with one of the $y$ got from the other term.

For instance try the same technique with the following:

$$\int y^{1000} \ln y \ \text{d}y$$

• Why do you divide through by d$y$? What does an integral mean if nothing says what it is respect to? – wnoise Jul 10 '11 at 2:04
• There are two types of student who leave out the $d?$, those who really know what they are doing and those who really don't. For the second kind, putting in the $d?$ is an indispensable guide to carrying out substitution, and integration by parts, without getting lost. – André Nicolas Jul 10 '11 at 8:38
• My \$0.02 based on teaching freshman calculus is that I think it's a bad idea for instructors to leave out the dx or dy in an integral when teaching beginning students. – idmercer Jul 10 '11 at 15:22 We go for integration by parts, since we are looking at the product of two unrelated functions, and there is no obvious substitution. I do not know what notation you use when setting up an integration by parts, so I will guess. The integration by parts "formula" can be written in two completely equivalent ways, apart from the choice of names. These are $$\int u\,dv=uv -\int v\,du\qquad\text{and}$$ $$\int v\,du=uv -\int u\,dv.$$ This formula comes directly from the Product Rule for derivatives. Please consult your text for details. In integration by parts, often the key question is: what shall I choose as$u$(and what shall I choose as$dv$). Maybe we should let$u=y$. Then$dv$must be$\ln y \,dy$. That sounds good, we will then have$du=dy$, which is a simplification. Great. We have let$dv=\ln y\,dy$, so we need to find$v$, that is, to integrate$\ln y$. Looks a little ugly, or at least not immediate. Let's not despair, maybe the integral of$\ln y$will not be too hard. But we might as well take a side glance at the other plausible choice,$du=y\,dy$,$v=\ln y$. Then$u=y^2/2$(we don't worry about constants of integration yet). Ouch, looks worse than$y$! And what looks worse usually is worse. But let's persist:$dv=(1/y)\,dy$. That's really good, ugly stuff is gone, we are basically finished. So, formally now, let$du=y\,dy$, let$v=\ln y$. Then$u=y^2/2$,$dv=(1/y)\,dy$. (Do lay it out more nicely than I have.) So we have $$\int y\ln y \,dy=uv-\int u\,dv=\frac{y^2}{2}\ln y-\int \frac{y}{2}dy.$$ Integrate, remembering now about the constant of integration. $$\int y\ln y\,dy=\frac{y^2\ln y}{2}-\frac{y^2}{4}+C.$$ Comment: When we integrate$x^3e^{2x}$, for example, the integration by parts strategy focuses on lowering the power of$x$. That is "usually" a good strategy. Exceptions to this general heuristic are things like$x^3\ln x$, and$x^3 \arctan x$. Extreme examples are$\int \ln x\,dx$,$\int \arctan x\,dx$, and$\int \arcsin x\, dx$, where it looks as if we don't even have a "product". But if, for example in$\int \ln x \,dx$, we let$du=dx$,$v=\ln x$, everything works out quickly. Added: Could substitution be a feasible approach? It seems sensible to try for$z=\ln y$, or equivalently$y=e^z$. Then$dy=e^z\,dz$. Our integral becomes $$\int ze^{2z}\,dz.$$ This is an integration by parts, so the substitution didn't buy us much! But it's a more "normal" integration by parts. The rest is routine. Here is a method of doing this by substitution. Put$y=e^{x}$. Then you have$dy = e^{x} \ dx. Substituting we have \begin{align*} \int e^{2x} \cdot x \ dx &= \frac{1}{4}\int e^{t} \cdot t\ dt \qquad \Bigl[ \text{substituting} \ t = 2x \Bigr] \\ &= \frac{1}{4}\int e^{t} \cdot \bigl( t + 1 \bigr) \ dt - \frac{1}{4}\int e^{t} \ dt \\ &= \frac{1}{4}e^{t} \cdot t -\frac{1}{4} e^{t} + C \qquad\qquad\qquad \Bigl[ \because \small \int e^{x} \cdot \Bigl( f(x) + f'(x) \Bigr) \ dx = e^{x} \cdot f(x) + C \ \Bigr] \\ &=\frac{1}{4}\cdot e^{2x} \cdot 2x - \frac{1}{4}\cdot e^{2x} +C \\ &=\log{y} \cdot y^{2} \times \frac{1}{2} - \frac{1}{4} \cdot y^{2} +C \end{align*} • +1 This seems pretty original. About notation: wouldn't\left(t+1\right)$be a little bit better? For the reason that some people might confuse it with fractional part en.wikipedia.org/wiki/Fractional_part I do not know whether it is a general convention, but when I was younger I was taught to use brackets from the most inner to the most outer in this order$\{[(\bullet)]\}$. – Martin Sleziak Jul 10 '11 at 6:30 • Well, aren't you disguising an integration by parts in the third step here? By the way: my comment to Aryabhata's answer was a (weak) joke alluding to this. – t.b. Jul 10 '11 at 7:06 • @Theo: Well, yes. But for a person who is aware of such formulas this may not take much time. After all working$e^{x}$is so fascinating. – user9413 Jul 10 '11 at 7:50 • @Martin: I think people coming here are matured enough to know when$\{$denotes a fractional part and when$\{\}$denotes a bracket operator. – user9413 Jul 10 '11 at 7:52 Here's another way to arrive at the result (which also provides a partial intuition). Fix$0 < y \leq 1\$. Then, $$\int_0^y {u\ln u \,du} = \int_0^y {u\bigg( - \int_u^1 {\frac{1}{v}} \,dv\bigg)du} = \int_{u = 0}^y {\bigg(\int_{v = u}^1 {\frac{{ - u}}{v}\,dv\bigg)du} } .$$ Interchanging the order of integration (as usual, it is helpful to draw a picture) yields $$\int_0^y {u\ln u \,du} = \int_{v = 0}^y {\bigg(\int_{u = 0}^v {\frac{{ - u}}{v}\,du\bigg)dv} } + \int_{v = y}^1 {\bigg(\int_{u = 0}^y {\frac{{ - u}}{v}\,du} \bigg)dv} .$$ Hence $$\int_0^y {u\ln u \,du} = \int_{v = 0}^y {\frac{1}{v}} \frac{{ - v^2 }}{2}\,dv + \int_{v = y}^1 {\frac{1}{v}\frac{{ - y^2 }}{2}\,dv} = - \frac{{y^2 }}{4} + \frac{{y^2 \ln y}}{2}.$$