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

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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

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#### Chris L T521

##### Well-known member
Staff member
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:
$$2\ln 2-4+\pi+2=\boxed{2\ln 2-2+\pi}$$

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