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Question Number 112487    Answers: 0   Comments: 0

solve ∫_0 ^∞ ((ln(1+x^3 ))/(1+x^2 ))dx

$${solve} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 112245    Answers: 0   Comments: 0

(1/(1+x^2 ))+(1/(1+x^4 ))+(1/(1+x^8 ))+..... ( x>1)

$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{8}} }+.....\:\:\:\:\:\:\:\:\:\:\left(\:{x}>\mathrm{1}\right) \\ $$

Question Number 112231    Answers: 0   Comments: 0

∫z^3 e^(−(4/3)z^3 ) dz

$$\int{z}^{\mathrm{3}} {e}^{−\frac{\mathrm{4}}{\mathrm{3}}{z}^{\mathrm{3}} } {dz} \\ $$

Question Number 112217    Answers: 0   Comments: 2

proporsed by m.n july 1970 evaluate Σ_(n=1) ^∞ (H_n /(n2^n )) solution recall thatΣ_(n=1) ^∞ H_n x^n =((ln(1−x))/(x−1)) ∵((H_n x^n )/n)=∫_0 ^x H_n t^(n−1) dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t(t−1)))dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t−1))dt−∫_0 ^x ((ln(1−t))/t)dt=A−B let A=∫_0 ^x ((ln(1−t))/(t−1))dt=−∫_0 ^x ((ln(1−t))/(1−t))dt A=−[−ln^2 (1−t)]_0 ^x +∫_0 ^x −((ln(1−t))/(1−t))dt 2A=ln^2 (1−x) A=(1/2)ln^2 (1−x)......(1) then B=∫_0 ^x ((ln(1−t))/t)dt=∫_0 ^1 ((ln(1−t))/t)dt+∫_1 ^x ((ln(1−t))/t)dt B=−Li_2 (1)−[Li_2 (x)−Li_2 (1)] B=−Li_2 (x)....(2) but I=A−B I=Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=(1/2)ln^2 (2)+Li_2 (x) ∵Σ_(n=1) ^∞ ((H_n x^n )/n)=(1/2)ln^2 (2)+Li_2 (x) but x=(1/2) Σ_(n=1) ^∞ (H_n /(n2^n ))=(1/2)ln^2 (2)+Li_2 ((1/2))=(1/2)ln^2 (2)+(π^2 /(12))−(1/2)ln^2 (2) ∵Σ_(n=1) ^∞ (H_n /(n2^n ))=(π^2 /(12)) by mathdave(06/09/2020)

$${proporsed}\:{by}\:{m}.{n}\:{july}\:\mathrm{1970} \\ $$$${evaluate} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} } \\ $$$${solution}\: \\ $$$${recall}\:{that}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} =\frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}−\mathrm{1}} \\ $$$$\because\frac{{H}_{{n}} {x}^{{n}} }{{n}}=\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}−\mathrm{1}}{dt}−\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}={A}−{B} \\ $$$${let}\:{A}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}−\mathrm{1}}{dt}=−\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$${A}=−\left[−\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)\right]_{\mathrm{0}} ^{{x}} +\int_{\mathrm{0}} ^{{x}} −\frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$\mathrm{2}{A}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)......\left(\mathrm{1}\right) \\ $$$${then}\: \\ $$$${B}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}+\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$${B}=−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)−\left[{Li}_{\mathrm{2}} \left({x}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\right] \\ $$$${B}=−{Li}_{\mathrm{2}} \left({x}\right)....\left(\mathrm{2}\right) \\ $$$${but}\:{I}={A}−{B} \\ $$$${I}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left({x}\right) \\ $$$$\because\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} {x}^{{n}} }{{n}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left({x}\right) \\ $$$${but}\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\because\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${by}\:{mathdave}\left(\mathrm{06}/\mathrm{09}/\mathrm{2020}\right) \\ $$

Question Number 112364    Answers: 1   Comments: 0

(1) Let z = (√(3+4i)) + (√(−3−4i)) , where square root is taken with positive real part. Then Re(z) is _ (a)3 (b) 4 (c) 2 (d) 1 (2)Suppose a,b,c are in AP and a^2 ,b^2 ,c^2 are in GP. If a<b<c and a+b+c=(3/2), then a = _ (a) (1/(2(√2))) (b) (1/(2(√3))) (c) (((√3)−2)/(2(√3))) (d) (((√2)−2)/(2(√2))) (3)A man sent 7 letters to his 7 friend . The letters are kept in addressed envelopes at random. The probability that 3 friend receive correct letters and 4 letters go to wrong destination is _ (a) (1/8) (b) (1/(16)) (c) (3/(32)) (d) (5/(64))

$$\left(\mathrm{1}\right)\:{Let}\:{z}\:=\:\sqrt{\mathrm{3}+\mathrm{4}{i}}\:+\:\sqrt{−\mathrm{3}−\mathrm{4}{i}}\:,\:{where} \\ $$$${square}\:{root}\:{is}\:{taken}\:{with}\:{positive}\: \\ $$$${real}\:{part}.\:{Then}\:{Re}\left({z}\right)\:{is}\:\_ \\ $$$$\left({a}\right)\mathrm{3}\:\:\:\:\:\left({b}\right)\:\mathrm{4}\:\:\:\:\:\left({c}\right)\:\mathrm{2}\:\:\:\:\:\left({d}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right){Suppose}\:{a},{b},{c}\:{are}\:{in}\:{AP}\:{and}\:{a}^{\mathrm{2}} ,{b}^{\mathrm{2}} ,{c}^{\mathrm{2}} \\ $$$${are}\:{in}\:{GP}.\:{If}\:{a}<{b}<{c}\:{and}\:{a}+{b}+{c}=\frac{\mathrm{3}}{\mathrm{2}}, \\ $$$${then}\:{a}\:=\:\_ \\ $$$$\left({a}\right)\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\left({c}\right)\:\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\left({d}\right)\:\frac{\sqrt{\mathrm{2}}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\left(\mathrm{3}\right){A}\:{man}\:{sent}\:\mathrm{7}\:{letters}\:{to}\:{his}\:\mathrm{7}\:{friend} \\ $$$$.\:{The}\:{letters}\:{are}\:{kept}\:{in}\:{addressed}\: \\ $$$${envelopes}\:{at}\:{random}.\:{The}\:{probability} \\ $$$${that}\:\mathrm{3}\:{friend}\:{receive}\:{correct}\:{letters} \\ $$$${and}\:\mathrm{4}\:{letters}\:{go}\:{to}\:{wrong}\:{destination} \\ $$$${is}\:\_ \\ $$$$\left({a}\right)\:\frac{\mathrm{1}}{\mathrm{8}}\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\left({c}\right)\:\frac{\mathrm{3}}{\mathrm{32}}\:\:\:\:\:\left({d}\right)\:\frac{\mathrm{5}}{\mathrm{64}} \\ $$

Question Number 112209    Answers: 0   Comments: 6

A triangle ABC has the following properties BC=1, AB=BC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\mathrm{AB}=\mathrm{BC}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

Question Number 112208    Answers: 1   Comments: 3

Question Number 112206    Answers: 0   Comments: 0

∫ln(sin(x)) dx

$$\int{ln}\left({sin}\left({x}\right)\right)\:{dx} \\ $$

Question Number 112199    Answers: 0   Comments: 0

Let Ω denote the circumcircle of ABC. The tangent to Ω at A meets BC at X. Let the angle bisectors of ∠AXB meet AC and AB at E and F respectively. D is the foot of the angle bisector from ∠BAC on BC. Let AD intersect EF at K and Ω again at L(other than A). Prove that AEDF is a rhombus and further prove that the circle defined by triangle KLX passes through the midpoint of line segment BC.

$$\mathrm{Let}\:\Omega\:\mathrm{denote}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{ABC}. \\ $$$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\Omega\:\mathrm{at}\:\mathrm{A}\:\mathrm{meets}\:\mathrm{BC}\:\mathrm{at}\:\mathrm{X}. \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{bisectors}\:\mathrm{of}\:\angle\mathrm{AXB}\:\mathrm{meet} \\ $$$$\mathrm{AC}\:\mathrm{and}\:\mathrm{AB}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{F} \\ $$$$\mathrm{respectively}.\:\mathrm{D}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{bisector}\:\mathrm{from}\:\angle\mathrm{BAC}\:\mathrm{on}\:\mathrm{BC}.\:\mathrm{Let}\:\mathrm{AD} \\ $$$$\mathrm{intersect}\:\mathrm{EF}\:\mathrm{at}\:\mathrm{K}\:\mathrm{and}\:\Omega\:\mathrm{again}\:\mathrm{at} \\ $$$$\mathrm{L}\left(\mathrm{other}\:\mathrm{than}\:\mathrm{A}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{AEDF}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rhombus}\:\mathrm{and}\:\mathrm{further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{triangle}\:\mathrm{KLX}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment} \\ $$$$\mathrm{BC}. \\ $$

Question Number 112195    Answers: 1   Comments: 0

Σ_(n=1 ) ^(11) (((−1)^(n+1) (4n+2))/(4n(n+1))) please help

$$\underset{\mathrm{n}=\mathrm{1}\:} {\overset{\mathrm{11}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \left(\mathrm{4n}+\mathrm{2}\right)}{\mathrm{4n}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$

Question Number 112189    Answers: 0   Comments: 0

prove that ∫_0 ^∞ (1/(cos(x)+sinh(x)))dx=1.4917.

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{cos}\left({x}\right)+{sinh}\left({x}\right)}{dx}=\mathrm{1}.\mathrm{4917}. \\ $$

Question Number 112179    Answers: 1   Comments: 1

Question Number 112173    Answers: 1   Comments: 0

find the solution of equation (√(cos 2x−sin^3 x+3)) = sin x

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation}\: \\ $$$$\sqrt{\mathrm{cos}\:\mathrm{2x}−\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{3}}\:=\:\mathrm{sin}\:\mathrm{x}\: \\ $$

Question Number 112169    Answers: 2   Comments: 0

∫ tan^3 x sec^3 x dx ?

$$\:\int\:\mathrm{tan}\:^{\mathrm{3}} {x}\:\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}\:? \\ $$

Question Number 112163    Answers: 1   Comments: 0

In a class of 30 pupils , 12 are girls and 2 of them are short−sighted. Among 18 boys , 6 are short−sighted . If a pupil is selected at random, what is the probability that the pupil chosen will be { (((a) a girls)),(((b) short−sighted)),(((c) a short−sighted boy)) :}

$$\mathrm{In}\:\mathrm{a}\:\mathrm{class}\:\mathrm{of}\:\mathrm{30}\:\mathrm{pupils}\:,\:\mathrm{12}\:\mathrm{are}\:\mathrm{girls}\:\mathrm{and}\: \\ $$$$\mathrm{2}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{short}−\mathrm{sighted}.\:\mathrm{Among}\: \\ $$$$\mathrm{18}\:\mathrm{boys}\:,\:\mathrm{6}\:\mathrm{are}\:\mathrm{short}−\mathrm{sighted}\:.\:\mathrm{If}\:\mathrm{a}\:\mathrm{pupil} \\ $$$$\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{pupil}\:\mathrm{chosen}\:\mathrm{will}\:\mathrm{be}\: \\ $$$$\begin{cases}{\left(\mathrm{a}\right)\:\mathrm{a}\:\mathrm{girls}}\\{\left(\mathrm{b}\right)\:\mathrm{short}−\mathrm{sighted}}\\{\left(\mathrm{c}\right)\:\mathrm{a}\:\mathrm{short}−\mathrm{sighted}\:\mathrm{boy}}\end{cases} \\ $$

Question Number 112162    Answers: 1   Comments: 3

Question Number 112159    Answers: 1   Comments: 0

(1)Find the equation of hyperbola with centre point at (1,−2) and coordinates of foci is (6,−2) and (−4,−2) (2) If hyperbola ((x^2 −2nx+n^2 )/(25))−((y^2 −2my+m^2 )/(16))=1 have a asympyotes passes through at (0,1), then 5m−4n =

$$\left(\mathrm{1}\right)\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{hyperbola}\:\mathrm{with}\: \\ $$$$\mathrm{centre}\:\mathrm{point}\:\mathrm{at}\:\left(\mathrm{1},−\mathrm{2}\right)\:\mathrm{and}\:\mathrm{coordinates} \\ $$$$\mathrm{of}\:\mathrm{foci}\:\mathrm{is}\:\left(\mathrm{6},−\mathrm{2}\right)\:\mathrm{and}\:\left(−\mathrm{4},−\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{If}\:\mathrm{hyperbola}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2nx}+\mathrm{n}^{\mathrm{2}} }{\mathrm{25}}−\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{2my}+\mathrm{m}^{\mathrm{2}} }{\mathrm{16}}=\mathrm{1} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{asympyotes}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{at}\: \\ $$$$\left(\mathrm{0},\mathrm{1}\right),\:\mathrm{then}\:\mathrm{5m}−\mathrm{4n}\:=\: \\ $$

Question Number 112145    Answers: 1   Comments: 0

4^(tan^2 x) + 2^(1/(cos^2 x)) − 80 = 0

$$\mathrm{4}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}} \:−\:\mathrm{80}\:=\:\mathrm{0} \\ $$

Question Number 112136    Answers: 2   Comments: 0

find (5/(6+7sin 2β)) , if tan β = 0.2

$${find}\:\:\:\frac{\mathrm{5}}{\mathrm{6}+\mathrm{7sin}\:\mathrm{2}\beta}\:,\:\:\:{if}\:\:\:\mathrm{tan}\:\beta\:\:=\:\:\mathrm{0}.\mathrm{2} \\ $$

Question Number 112133    Answers: 1   Comments: 0

Question Number 112128    Answers: 5   Comments: 2

Question Number 112318    Answers: 2   Comments: 2

A triangle ABC has the following properties BC=1, AB=AC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

Question Number 112135    Answers: 3   Comments: 0

solve ∫_(−1) ^1 ∣3^x −2^x ∣dx

$${solve} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \mid\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mid{dx} \\ $$

Question Number 112119    Answers: 1   Comments: 0

....calculus.... prove that::: if Ω =∫_(0 ) ^( 1) ln(ln(1−(√x) ))dx then Re(Ω) := −γ + ln(2).... m.n. july 1970#

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{calculus}.... \\ $$$${prove}\:{that}::: \\ $$$${if}\:\:\:\Omega\:=\int_{\mathrm{0}\:\:} ^{\:\mathrm{1}} {ln}\left({ln}\left(\mathrm{1}−\sqrt{{x}}\:\right)\right){dx} \\ $$$${then} \\ $$$$\mathscr{R}{e}\left(\Omega\right)\::=\:−\gamma\:+\:{ln}\left(\mathrm{2}\right).... \\ $$$$ \\ $$$${m}.{n}.\:{july}\:\mathrm{1970}# \\ $$

Question Number 112114    Answers: 0   Comments: 0

solution of Φ=∫_0 ^1 xH_x dx =^(γ+ψ(x+1)) ∫_0 ^( 1) x(γ+ψ(x+1))dx =^(ψ(x+1)=(1/x)+ψ(x)) ∫_0 ^( 1) x(γ+(1/x)+ψ(x))dx =(γ/2)+1+ ∫_0 ^( 1) x.(d/dx)(ln(Γ(x))) =(γ/2)+1+[xln(Γ(x))]_0 ^1 −∫_0 ^( 1) ln(Γ(x))dx we know (why?) ∫_0 ^( 1) ln(Γ(x))dx=ln((√(2π)) ) and lim_(x→0^+ ) (xln(Γ(x)))=0 (why??) finally Φ =∫_0 ^( 1) xH_x dx=(γ/2)+1−ln((√(2π))) ✓ m.n.july 1970....

$$\:\:{solution}\:{of} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} {xH}_{{x}} {dx}\:\overset{\gamma+\psi\left({x}+\mathrm{1}\right)} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\psi\left({x}+\mathrm{1}\right)\right){dx}\: \\ $$$$\:\overset{\psi\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)\right){dx} \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\frac{{d}}{{dx}}\left({ln}\left(\Gamma\left({x}\right)\right)\right) \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\left[{xln}\left(\Gamma\left({x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx} \\ $$$$\:\:{we}\:\:{know}\:\left({why}?\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}={ln}\left(\sqrt{\mathrm{2}\pi}\:\right) \\ $$$$\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left({xln}\left(\Gamma\left({x}\right)\right)\right)=\mathrm{0}\:\left({why}??\right) \\ $$$${finally}\:\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}=\frac{\gamma}{\mathrm{2}}+\mathrm{1}−{ln}\left(\sqrt{\mathrm{2}\pi}\right)\:\checkmark \\ $$$$\:\:{m}.{n}.{july}\:\mathrm{1970}.... \\ $$$$\: \\ $$

Question Number 112096    Answers: 3   Comments: 0

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