Welcome to our community

Be a part of something great, join today!

Problem of the Week #101 - March 3rd, 2014

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Evaluate $\displaystyle\lim_{x\to 0^{+}}\left(\int_{x}^{1}\dfrac{\ln{(1+t)}}{\sqrt{t}}dt+\int_{0}^{x}\dfrac{\sin{2t}}{\sqrt{4+t^2}\int_{0}^{x}(\sqrt{y+1}-1)dy}dt\right)$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
This week's problem was correctly answered by Opalg and Pranav. You can find Pranav's solution below.

The given limit can be written as:
$$\int_0^1 \frac{\ln(1+t)}{\sqrt{t}}\,dt + \lim_{x\rightarrow 0^+} \cfrac{\int_0^x \frac{\sin(2t)}{\sqrt{4+t^2}}\,dt}{\int_0^x (\sqrt{y+1}-1)\,dy}$$
I evaluate the above separately.

First I evaluate the following integral:
$$\int_0^1 \frac{\ln(1+x)}{\sqrt{t}}\,dt$$
Substitute $t=x^2 \Rightarrow dt=2x\,dx$, the integral becomes
$$2\int_0^1 \ln(1+x^2)\,dx=2\left(\left(\ln(1+x^2)\cdot x\right|_0^1-\int_0^1 \frac{2x^2}{1+x^2}\,dx\right)$$
(I used integration by parts)
$$=2\left(\ln 2-2\left(\int_0^1 1-\frac{1}{1+x^2} \,\,dx\right)\right)=2\left(\ln 2 -2\left(1-\frac{\pi}{4}\right)\right)=\boxed{2\ln 2-4+\pi}$$

Next, I evaluate the following limit,
$$\lim_{x\rightarrow 0^+} \cfrac{\int_0^x \frac{\sin(2t)}{\sqrt{4+t^2}}\,dt}{\int_0^x (\sqrt{y+1}-1)\,dy}$$
Substituting $x=0$ gives $\frac{0}{0}$ so L'Hopital's rule can be used i.e
$$\lim_{x\rightarrow 0^+} \cfrac{\int_0^x \frac{\sin(2t)}{\sqrt{4+t^2}}\,dt}{\int_0^x (\sqrt{y+1}-1)\,dy}=\lim_{x\rightarrow 0^+} \cfrac{\frac{\sin 2x}{\sqrt{4+x^2}}}{\sqrt{x+1}-1}=\lim_{x\rightarrow 0^+} \frac{\sin 2x}{\sqrt{4+x^2}(\sqrt{x+1}-1)}=\lim_{x\rightarrow 0^+} \frac{\sin (2x) (\sqrt{x+1}+1)}{x\sqrt{4+x^2}}$$
$$=\lim_{x\rightarrow 0^+} 2\frac{\sin (2x) (\sqrt{x+1}+1)}{2x\sqrt{4+x^2}}$$
The limit of $\sin(2x)/(2x)$ as $x$ tends to zero is 1, hence, the above limit is equal to $\boxed{2}$.

Putting everything together, the final result is:
$$2\ln 2-4+\pi+2=\boxed{2\ln 2-2+\pi}$$
 
Status
Not open for further replies.