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.

  • $\begingroup$ Before speaking of "the" naive method, maybe you should consider that there might be many naive methods. How about this one: Circumscribe a right circular cone about a sphere. Look at the right circular cone with two nappes whose axis coincides with that of the cylinder. Look at cross-sections: intersections of this whole structure with planes orthogonal to the axis. Show that the disk where that plane intersects the sphere has the same area as the ring in which that plane intersects the region that is$\,\ldots\qquad$ upload.wikimedia.org/wikipedia/commons/7/72/DoubleCone.png $\endgroup$ – Michael Hardy Apr 14 '16 at 22:24
  • $\begingroup$ $\ldots\,$outside of the cone and inside of the cylinder. Conclude that the volume enclosed by the sphere is the same as the volume of the region inside the cylinder but outside the cone. From this you get the volume of the sphere: $\frac 4 3 \pi r^3$. $\qquad$ $\endgroup$ – Michael Hardy Apr 14 '16 at 22:25
  • $\begingroup$ Also, the "naive" method isn't really particularly naive in general. $\qquad$ $\endgroup$ – Michael Hardy Apr 14 '16 at 22:26
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    $\begingroup$ @MichaelHardy, this might be one of my favorite comments ever. Also, your "naive" method is a beautiful way of computing the volume of the sphere (and doesn't strike me as naive at all). I'm bookmarking this question so I always know where to find this construction. $\endgroup$ – Nicholas Stull Apr 15 '16 at 0:55
  • $\begingroup$ @NicholasStull : I'm glad you liked it. This is the method of finding the volume of a sphere that I saw in a geometry course in 11th grade. The the volume of a cone is $1/3 \times \text{base} \times \text{height}$ was proved by a similar method: Look at the cone and at a triangular pyramid with the same base-area and the same height, and observe that their corresponding horizontal cross-sections have equal areas; conclude that they have equal volumes. The fact that the area of the triangular pyramid is${}\,\ldots\qquad$ $\endgroup$ – Michael Hardy Apr 15 '16 at 15:47

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 $$ \int_1^x\ln t\,dt=x\ln x-x+1 $$


An interesting approach (changing the bounds for simplicity):

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


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