Evaluating a certain integral without the fundamental theorem

Question

I'm a TA for a calculus course. And they recently began calculating definite integrals using a definition equivalent to Riemann's criterion. Of course, the type of things they were calculating were fairly elementary such as $$\int_0^1x\;dx\qquad\text{and}\qquad\int_0^1x^2\;dx$$ Knowing full well that the fundamental theorem of calculus was on the itinerary, I decided to give them an appreciation for the result by proving a much more general result (still using rectangles). Namely I showed: $$\int_a^b x^n\;dx=\frac{b^{n+1}}{n+1}-\frac{a^{n+1}}{n+1}$$ This can be calculated in a way that parallels the calculation of the above examples. As long as one knows that $$\lim_{m\rightarrow\infty}\frac{1^n+2^n+\cdots+m^n}{\frac{m^{n+1}}{n+1}}=1$$ one is able to proceed. Granted, I had to give a loose argument for why this is true, but knowing that $$1+2+\cdots+n=\frac{1}{2}n^2+\cdots \qquad\text{ and } 1^2+2^2+ \cdots +n^2 = \frac{1}{3}n^3 + \cdots$$ The pattern seems plausible. I thought this was cute, so I also gave them the proof that $$\int_0^x\cos t\;dt=\sin x$$ which can be derived with rectangles using Dirichlet's identity: $$1+2\sum_{k=1}^n\cos(kx)=\frac{\sin\left([n+1/2]x\right)}{\sin(x/2)}$$ To be sure, many students found this un-amusing, but they all greatly affirmed that they were glad to have the fundamental theorem after it was delivered to them. So goal achieved. But I was intrigued by how many other integrals could I evaluate using the naive method? $$\int_0^x e^t\;dt$$ isn't too bad as it's a geometric sum. The next thing in line was, of course, $$\int_1^x\ln t\; dt$$ This is where I ran into trouble. I had been using the fact that $$\int_a^b f(x)\;dx=\lim_{n\rightarrow\infty}\sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right)\frac{b-a}{n}$$ for integrable $f$ to do the fore-going facts. But this approach seems intractable for $$\int_1^x\ln t\; dt$$ At least, I don't have the requisite limit knowledge or 'algebraic trick' needed to proceed. I was able to calculate this with the fact that $$\int_0^{\ln x}e^t\;dt+\int_1^x\ln t\;dt=x\ln x$$ which is a relationship that can be proven naively. But I was hoping someone here knew the 'trick' needed to calculate $$\int_1^x \ln t\;dt$$ without the fundamental theorem or relying on the insight to reflect the area in question. Any help is appreciated.

Answer 1

For logarithmic integrals a subdivision into geometric progression is often convenient. Set $r=\sqrt[n]{x}$ and consider the upper sum $$ \sum_{k=1}^n (r^k-r^{k-1})\ln(r^k)= \ln r\sum_{k=1}^n k(r^k-r^{k-1}) $$ It's easy to show, by induction, that $$ \sum_{k=1}^n k(r^k-r^{k-1})=nr^n-\sum_{k=0}^{n-1}r^k =nr^n-\frac{r^n-1}{r-1} $$ Putting back $r=x^{1/n}$, we get, for the upper sum, the expression $$ \left(x-\frac{x-1}{n(x^{1/n}-1)}\right)\ln x $$ Now, $$ \lim_{n\to\infty}n(x^{1/n}-1)=\lim_{t\to0^+}\frac{x^t-1}{t}=\ln x $$ so the limit of the upper sums is $$ \left(x-\frac{x-1}{\ln x}\right)\ln x=x\ln x-x+1 $$

Check similarly for the lower sums and see that this agrees with $$ \int_1^x\ln t\,dt=x\ln x-x+1 $$

Answer 2

An interesting approach (changing the bounds for simplicity):

$$\int_0^x\ln(t)~\mathrm dt=\lim_{s\to0}\frac{\mathrm d}{\mathrm ds}\int_0^xt^s~\mathrm dt=\lim_{s\to0}\frac{\mathrm d}{\mathrm ds}\frac{x^{s+1}}{s+1}=\lim_{s\to0}\frac{\ln(x)-1}{(s+1)^2}x^{s+1}=x\ln(x)-x$$