Calculus Marathon

[warut]

ข้อความเดิมของคุณ Mastermander:
38.Evaluate

$$\int_0^1\frac{\arctan x}{x}\ dx$$

ข้อความเดิมของคุณ nooonuii:
ข้อ 38 นี่ ไม่โหดไปหน่อยเหรอครับ ผมลองให้น้องเปิ้ล( Maple ) คิดดูแล้วได้คำตอบออกมาเป็น Catalan constant อยากทราบเฉลยข้อนี้จังเลยครับ

To prove, just expand $\tan^{-1}x$ into its Maclaurin series, and then integrate term by term. The result you get is the series that is used to define Catalan's constant.

[Mastermander]

40.

$$\int_{-\infty}^\infty\frac{\cos 3x}{(x^2+1)^2}\ dx$$

[nooonuii]

By mimicking the proof of problem#37 with $\displaystyle{ f(z) = \frac{e^{3iz}}{(1+z^2)^2} }$, we can show that
$$\int_{-\infty}^\infty\frac{\cos 3x}{(x^2+1)^2}\ dx = 2\pi i\text{Res}(f(z),i) = \frac{2\pi}{e^3}.$$

[Mastermander]

41. Evaluate

$$\int_0^{\infty} t^2e^{-3t}\sinh(t) \ dt$$

[nooonuii]

39. ดูเหมือนจะมีคำตอบสวยๆ แต่คิดยังไงก็ไม่ออกครับ ยอมแพ้
41. ไม่ยากครับ ใช้ tabular integration by parts จะได้

$$\displaystyle{\int_0^{\infty} t^2e^{-3t}\sinh(t) \ dt = \frac{1}{2}\int_0^{\infty} t^2(e^{-2t} - e^{-4t}) \ dt = \frac{7}{64}}$$

[Mastermander]

ผมมีแต่คำตอบครับ ไม่เคยเห็นใครเผย Solution สักที

[nooonuii]

42. จงหาค่าของ $$\lim_{x\rightarrow\infty} (\sqrt[3]{ x^3+x^2 } - \sqrt[3]{ x^3-x^2 }) $$

[Mastermander]

Let $a=x^3+x^2,b=x^3-x^2$

$$\sqrt[3]{a}-\sqrt[3]{b}=(\sqrt[3]{a}-\sqrt[3]{b})(\frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}})$$

$$=\frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}=\frac{2x^2}{(x^3+x^2)^{2/3}+(x^6-x^4)^{1/3}+(x^3-x^2)^{2/3}}$$

divided both by $x^2$

$$ =\frac{2}{(1+x^{-1})^{2/3}+(1+x^{-2})^{1/3}+(1-x^{-1})^{2/3}} $$

take limit as x to ฅ , Hence limit $=\frac{2}{3}$

[Mastermander]

43.

[nooonuii]

Let $$a_n = \frac{1}{n}\sqrt[n]{(n+1)(n+2)\dots (n+n)} = \sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})\dots (1+\frac{n}{n}).}$$
Let $f(x) = \ln{(1+x)}$. Then $f$ is Riemann integrable on $[0,1]$ and $\displaystyle{ \int_0^1 f(x) dx = 2\ln{2}-1. }$
For each $n\in N,$ define a partition $\displaystyle{ P_n=\{0,\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n}\} } $.
Then $U(f,P_n) = \ln{a_n}$.
Since $f$ is Riemann integrable on $[0,1]$, $\displaystyle{ \int_0^1 f(x) dx = \lim_{n\rightarrow\infty} U(f,P_n)= \lim_{n\rightarrow\infty} \ln{a_n}. }$
Therefore, $\displaystyle{ \lim_{n\rightarrow\infty} \ln{a_n} = 2\ln{2} - 1 }$, i.e.,
$$\displaystyle{ \lim_{n\rightarrow\infty} a_n = \frac{4}{e}. }$$

[Mastermander]

44.

$$\int_0^1 \bigg(\ln\frac{1}{x}\bigg)^n \ dx$$

[nooonuii]

44. By substituting $t = -\ln{x}$, we have
$$\displaystyle{ \int_0^1 (-\ln{x})^n dx = \int_0^{\infty} t^ne^{-t}dt}$$
Let $\displaystyle{ I_n = \int_0^{\infty} t^ne^{-t}dt} $ for $n\in N.$
Integration by parts gives $I_n = nI_{n-1}.$
It is easy to see that $I_1 = 1.$
Thus $I_n=n!.$
Therefore, $$\displaystyle{ \int_0^1 (-\ln{x})^n dx = n!. }$$

[Mastermander]

45.

[Mastermander]

46.

[nooonuii]

45. ข้อนี้ถ้าผมไม่มี solution อยู่ในมือคงทำไม่ได้แน่ๆ

For $x>0$, we have
$$\displaystyle{\int_x^{x+1} \sin{t^2} dt = \frac{-\cos{(x+1)^2}}{2(x+1)}+\frac{\cos{x^2}}{2x}-\frac{1}{2}\int_x^{x+1}\frac{\cos{t^2}}{t^2} dt}$$
from integration by parts by picking $ \displaystyle{ u = \frac{1}{2t}, dv = 2t\sin{t^2}dt.}$
Thus $\displaystyle{ |\int_x^{x+1}\sin{t^2} dt | \leq \frac{|\cos{(x+1)^2|}}{2(x+1)}+\frac{|\cos{x^2}|}{2x}+\frac{1}{2}\int_x^{x+1}\frac{|\cos{t^2}|}{t^2} dt \leq \frac{1}{2(x+1)}+\frac{1}{2x}+\frac{1}{2}\int_x^{x+1}\frac{1}{t^2} dt = \frac{1}{x}.}$
Hence $\displaystyle{ |x^{1-\epsilon}\int_x^{x+1}\sin{t^2} dt | \leq \frac{1}{x^{\epsilon}} \rightarrow 0}$ as $x\rightarrow\infty .$
Therefore, $\displaystyle{ \lim_{x\rightarrow\infty}x^{1-\epsilon}\int_x^{x+1}\sin{t^2} dt = 0.}$


ส่วนข้อ 46 ตอบ 0.5 ไม่ยากถ้ารู้ว่าหน้าตาของอนุกรมทั้งเศษและส่วนเป็นอะไร