3
$\begingroup$

How can we evaluate the following integral:

$$\int_0^{1/10}\sum_{k=0}^9 \frac{1}{\sqrt{1+(x+\frac{k}{10})^2}}dx$$


I know basically how to calculate by using the substitution $x=\tan{\theta}$ :
$$\int_0^1 \frac{dx}{\sqrt{1+x^2}}$$
But I cannot find out a way to apply the result to the question.

$\endgroup$
  • $\begingroup$ Make the substitution $x+\frac{k}{10}=\tan(\theta)$ $\endgroup$ – JasonM Jun 12 '16 at 6:33
  • $\begingroup$ @JasonM That means we have to evaluate this ten times? Is there a better way? $\endgroup$ – The Integral Jun 12 '16 at 6:35
  • $\begingroup$ It's a finite sum, so you can swap the sum and integral's positions. $\endgroup$ – JasonM Jun 12 '16 at 6:36
  • $\begingroup$ @JasonM - this substitution would have helped if the square root sign is not present. $\endgroup$ – hypergeometric Jun 13 '16 at 15:50
  • $\begingroup$ @hypergeometric You guys are right, I tried it myself and it got messy. $\endgroup$ – JasonM Jun 13 '16 at 19:12
7
$\begingroup$

There is a better way.

We shall prove that:

$$\int_0^s f(x+ks)dx=\int_{ks}^{(k+1)s}f(x)dx \tag1$$

And hence:

$$\int_0^s[f(x)+f(x+s)+...+f(x+(n-1)s)]dx=\int_0^{ns}f(x)dx \tag2$$

Proof:

Using substitution $t\mapsto x+ks$,
$$\int_0^s f(x+ks)dx=\int_{ks}^{(k+1)s}f(t)dt$$
\begin{align} & \int_0^s[f(x)+f(x+s)+\cdots+f(x+(n-1)s)]dx \\ & = \int_0^sf(x)dx+\int_0^sf(x+s)dx+\cdots+\int_0^sf(x+(n-1)s)dx \\ & = \int_0^sf(x)dx+\int_s^{2s}f(x)dx+\cdots+\int_{(n-1)s}^{ns}f(x)dx \\ & = \int_0^{ns}f(x)dx \end{align}


Using the aforementioned results, your integral just becomes:
$$\int_0^1\frac{dx}{\sqrt{1+x^2}} \tag3$$
, which is exactly equal to your given integral!

$\endgroup$
4
$\begingroup$

Consider $$I_k=\int \frac{dx}{\sqrt{1+(x+\frac{k}{10})^2}}$$ and let $x+\frac{k}{10}=\sinh(y)$, $dx=\cosh(y)\, dy$ which make $$I_k=\int dy=y=\sinh^{-1} \left(x+\frac k{10}\right)$$ So,$$J_k=\int_0^{\frac 1{10}} \frac{dx}{\sqrt{1+(x+\frac{k}{10})^2}}=\sinh ^{-1}\left(\frac{k+1}{10}\right)-\sinh ^{-1}\left(\frac{k}{10}\right)$$ which leads to nicely telescoping terms.

I am sure that you can take it from here.

$\endgroup$
0
$\begingroup$

Put $x+\frac k{10}=\sinh u$ (i.e. $dx=\cosh u$), we have $$\begin{align} &{\large\int}_0^{1/10}\sum_{k=0}^9\frac 1{\sqrt{1+\left(x+\frac k{10}\right)^2}}dx\\ &=\sum_{k=0}^9{\large\int}_0^{1/10}\frac 1{\sqrt{1+\left(x+\frac k{10}\right)^2}}dx\\ &=\sum_{k=0}^9{\large\int}_\alpha^\beta\frac 1{\sqrt{1+\sinh^2u}}\;\cosh u\; du &&\scriptsize \text{where }\alpha=\sinh^{-1}\frac k{10}, \beta=\sinh^{-1}\frac{k+1}{10}\\ &=\sum_{k=0}^9{\large\int}_\alpha^\beta 1\; du\\ &=\sum_{k=0}^9\; \sinh^{-1}\left(\frac{k+1}{10}\right)-\sinh^{-1}\left(\frac k{10}\right)\\ &=\sinh^{-1}1-\sinh^{-1}0&&\scriptsize\text{by telescoping}\\\\ &=\color{red}{\ln(1+\sqrt{2})}\qquad\blacksquare \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.