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From the definition of a definite integral as the limit of a sum, evaluate $$\int_a^b\frac{1}{\sqrt x}dx$$


My progress so far:

$\begin{align} \int_a^b\frac{1}{\sqrt x}dx & =\lim_{n\to\infty}h\times\bigg(f(a)+f(a+h)+\cdots+f\big(a+(n-1)h\big)\bigg) \\ & = \lim_{n\to\infty}h\times\bigg(\frac1{\sqrt a}+\frac1{\sqrt {a+h}} +\cdots+\frac1{\sqrt {a+(n-1)h}}\bigg) \end{align}$


After this point, I cant't think of a way to combine the reciprocals of square roots into a neat form. Usually a convenient summation formula is applicable but how would I proceed in this case?

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  • $\begingroup$ Are you sure you're not given some sort of constants $a,b$? I don't recall ever seeing one without them in my school work, seems like it might make things a little too difficult at first glance. $\endgroup$ – Eevee Trainer Mar 29 at 2:45
  • $\begingroup$ @EeveeTrainer Yes, all the other questions in the exercise also have limits $a$,$b$. Using $nh=b-a$ still works even if you don't know what $a$ and $b$ are. $\endgroup$ – s0ulr3aper07 Mar 29 at 2:49
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We assume $0\leq a\leq b$. In order to cope with the square root function $\frac{1}{\sqrt{x}}$ it is convenient to use variable length intervals with length $j^2\frac{b-a}{n^2}$. When taking square roots we can factor out $j$ and summation is expected to become simpler. The corresponding Riemann sum is \begin{align*} \int_{a}^{b}\frac{1}{\sqrt{x}}\,dx=\lim_{n\to \infty}\sum_{j=1}^n\frac{1}{\sqrt{\color{blue}{a}+ j^2\frac{b-a}{n^2}}}\cdot\left(j^2\frac{b-a}{n^2}-(j-1)^2\frac{b-a}{n^2}\right)\tag{1} \end{align*} But it's still not easy due to the constant $\color{blue}{a}$. To overcome this difficulty we write the integral as difference of two improper integrals \begin{align*} \int_{a}^{b}\frac{1}{\sqrt{x}}\,dx=\int_{0}^{b}\frac{1}{\sqrt{x}}\,dx-\int_{0}^{a}\frac{1}{\sqrt{x}}\,dx\tag{2} \end{align*} If the limit of the Riemann sums of the improper integrals exists we have found the wanted Riemann integral and we are done.

We subdivide the interval $[0,b]$ in \begin{align*} [0,b]=\bigcup_{j=1}^n\left[(j-1)^2\frac{b}{n^2},j^2\frac{b}{n^2}\right] \end{align*} with length \begin{align*} j^2\frac{b}{n^2}-(j-1)^2\frac{b}{n^2}=\frac{(2j-1)b}{n^2}\qquad\qquad1\leq j\leq n \end{align*} and obtain the Riemann sum \begin{align*} \color{blue}{\int_{0}^b\frac{1}{\sqrt{x}}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\sqrt{\frac{j^2b}{n^2}}}\cdot\frac{(2j-1)b}{n^2}\tag{3}\\ &=\sqrt{b}\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^n\left(2-\frac{1}{j}\right)\tag{4}\\ &=\sqrt{b}\lim_{n\to\infty}\frac{1}{n}\left(2n-H_n\right)\tag{5}\\ &\,\,\color{blue}{=2\sqrt{b}}\tag{6} \end{align*}

We finally conclude from (2) and (6) \begin{align*} \color{blue}{\int_{a}^b\frac{1}{\sqrt{x}}\,dx=2\left(\sqrt{b}-\sqrt{a}\right)} \end{align*}

Comment:

  • In (3) we have the now convenient representation (1) evaluated at $a=0$.

  • In (4) we do some simplifications and factor out terms not dependent on $n$ resp. $j$.

  • In (5) we note the Harmonic numbers $H_n=\sum_{j=1}^n \frac{1}{j}$ grow asymptotically with $\log n$, so that \begin{align*} \lim_{n\to\infty}\frac{H_n}{n}=\lim_{n\to\infty}\frac{\log n}{n}=0. \end{align*}

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  • $\begingroup$ Thanks for the beautiful solution. However, while using $log \ n$ to approximate $H_n=\sum_{j=1}^n \frac{1}{j}$, there is difference of exactly $\gamma = 0.57721\ldots$ between the two when $\lim_{n\to \infty}$. So maybe adding the constant value would make this already brilliant answer even better? $\endgroup$ – s0ulr3aper07 Mar 29 at 18:24
  • $\begingroup$ Also, $j^2\frac{b}{n^2}-(j-1)^2\frac{b}{n^2}=\frac{(2j-1)b}{n^2}$ is I believe what you intended to write. $\endgroup$ – s0ulr3aper07 Mar 29 at 18:26
  • $\begingroup$ @s0ulr3aper07: You're welcome and many thanks for pointing to the mistakes. $\endgroup$ – Markus Scheuer Mar 29 at 18:39
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    $\begingroup$ @s0ulr3aper07: With respect to the harmonic numbers I think it's sufficient to use the dominant term $\log n$ of the asymptotic expansion, since we are dividing by $n$ when taking the limit. $\endgroup$ – Markus Scheuer Mar 29 at 18:40
  • $\begingroup$ I tried a similar approach for $\int_a^b{\frac1{x^2}dx}$ with variable length intervals of length $\sqrt{j} \frac{b-a}{\sqrt{n}}$ but unlike the resulting calculations above, these new calculations are not as clean. Specially because I get $\sum_{j=1}^n \frac{1}{j}(\sqrt{j}-\sqrt{j-1}$). Could you provide some help there please? $\endgroup$ – s0ulr3aper07 Mar 30 at 8:09

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