Brilliant Junior Integration Contest - Season 1

Lemma 1 : \(\displaystyle \sum_{r=1}^{n} H_{r} = (n+1)H_{n} - n\)

Proof : \(\displaystyle \sum_{r=1}^{n} H_{r} = \int_{0}^{1} \sum_{r=1}^{n} \dfrac{x^r-1}{x-1} \mathrm{d}x\)

\(\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x - nx + n}{(x-1)^2} \mathrm{d}x \)

\(\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x}{(x-1)^2} \mathrm{d}x - \int_{0}^{1} \dfrac{n}{x-1} \mathrm{d}x\)

\(\displaystyle = \text{A} + \text{B}\)

Using Integration By Parts on the first integral \(\text{A}\), we have,

\(\displaystyle \text{A} = -\left[\dfrac{x^{n+1}-x}{x-1}\right]_{0}^{1} + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x \)

\(\displaystyle = -n + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x\)

\(\displaystyle \implies \text{A} +\text{B} = -n + (n+1) \cdot \int_{0}^{1}\dfrac{x^n - 1}{x-1} \mathrm{d}x \)

\(\displaystyle = (n+1)H_{n} - n \)

Lemma 2 : \(\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \mathrm{d}x = \ln (n +1)\)

Proof : \(\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \ \mathrm{d}x = \int_{0}^{1} \int_{0}^{n} x^y \ \mathrm{d}y \ \mathrm{d}x\)

\(\displaystyle = \int_{0}^{n} \int_{0}^{1} x^y \ \mathrm{d}x \ \mathrm{d}y\)

\(\displaystyle = \int_{0}^{n} \dfrac{1}{y+1} \ \mathrm{d}y \)

\(\displaystyle = \ln (n+1)\)

Lemma 3 : \(\displaystyle \int_{0}^{1} \left(\dfrac{1}{\ln (x)} + \dfrac{1}{1-x}\right) \mathrm{d}x = \gamma\)

Proof : \(\displaystyle \int_{0}^{1} \left( \dfrac{1}{1-x} + \dfrac{1}{\ln (x)}\right) \mathrm{d}x = \lim_{n \to \infty} \int_0^1\left(\dfrac{1-t^n}{1-t}-\frac{t^{n-1}-1}{\ln t} \mathrm{d}t \right)\)

\(\displaystyle = \lim _{n \to \infty} (H_{n} - \ln n)\)

\(\displaystyle = \gamma\)

Now,

\(\displaystyle \text{I} = \int_{0}^{1} \left(\dfrac{1}{\ln x} + \dfrac{1}{1-x} - \dfrac{1}{2}\right)\dfrac{\mathrm{d}x}{x-1} \)

\(\displaystyle = \lim_{n \to \infty} \sum_{r=1}^{n} \int_{0}^{1} \left(\dfrac{x^{r-1}}{\ln x} + \dfrac{x^{r-1}}{1-x} - \dfrac{x^{r-1}}{2} \right)\)

Using Lemma 2 and Lemma 3 , we have,

\(\displaystyle \text{I} = \lim_{n \to \infty} \sum_{r=1}^{n} \left(\gamma + \ln r - H_{r-1} -\dfrac{1}{2r} \right) \)

\(\displaystyle = \lim_{n \to \infty} \left( n\gamma + \ln (n!) - \sum_{r=1}^{n} H_{r-1} - \dfrac{H_{n}}{2} \right)\)

Using Lemma 1,

\(\displaystyle \text{I} = \lim_{n \to \infty} \left( n\gamma + \ln (n!) -n H_{n-1} + n - 1 -\dfrac{H_{n}}{2} \right) \)

Using Stirling's Approximation, we have,

\(\displaystyle \text{I} = \lim_{n \to \infty} \left( n\gamma + \dfrac{1}{2} \ln (2\pi) - n +\dfrac{1}{2} \ln (n) + n\ln (n) -n H_{n-1} + n - 1 -\dfrac{H_{n}}{2} \right)\)

Simplifying using \(\displaystyle \lim_{n \to \infty} (H_{n} - \ln n) = \gamma\), we have,

\(\displaystyle \text{I} = \boxed{\dfrac{1}{2} \ln (2\pi) -\dfrac{1}{2} \gamma - \dfrac{1}{2}}\)

Solution to Problem 2

\[I = \int _{ 0 }^{ 1 }{ \frac { \arctan { \left( \sqrt { { x }^{ 2 }+2 } \right) } }{ \sqrt { { x }^{ 2 }+2 } \left( { x }^{ 2 }+1 \right) } \ dx } =\frac { 5{ \pi }^{ 2 } }{ 96 } \]

Lets split \(I\) as \(A \text{ and } B\) by using the property that \(\arctan(x) = \pi /2 -\arctan(1/x)\)

Thus our integral becomes \(I = A-B\),

where \(A = \displaystyle \pi /2\int_{0}^{1}\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})} \)

and \(B= \displaystyle \int_{0}^{1}\dfrac{\arctan(\frac{1}{\sqrt{x^{2} + 2}})}{(1+x^{2})(\sqrt{2+x^{2}})} \)

Evaluating A,

Let's evaluate A, without the limits,

\(\displaystyle \pi /2\int\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})} = \pi /2\arctan \dfrac{x}{\sqrt{x^{2} + 2}}\)

Which on substituting limits gives A = \(\pi ^{2}/12\)

Evaluating B,

Using \(1/a \arctan(1/a) = \int_{0}^{a} \frac{dy}{y^{2}+a^{2}}\),

Let \(a = \sqrt{x^2+2}\).

We get \(B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})} \)

\(B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})} - \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})} \)

\(2B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})}\)

\(2B = \displaystyle (\int_{0}^{1}\dfrac{dx}{1+x^{2}})^{2}\)

Thus \(B = \pi^{2} / 32\)

Hence \(I = A-B=\dfrac{5\pi^{2}}{96}\)

Generalization

Solution to Problem 1: \[{ I }_{ 1 }=\int _{ -\infty }^{ \infty }{ { \left( x+1 \right) }^{ 2 }{ e }^{ { -\left( x+1 \right) }^{ 2 } } } dx\]

Substitute: \({ \left( x+1 \right) }^{ 2 }=t\)

Therefore, we get: \[{ I }_{ 1 }=\frac { 1 }{ 2 }\int _{ -\infty }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt\]

This is symmetric about \(x=0\). Therefore, we can write it as: \[{ I }_{ 1 }=\int _{ 0 }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt\]

\[{ I }_{ 1 }=\Gamma \left( \frac { 3 }{ 2 } \right) =\frac { 1 }{ 2 } \Gamma \left( \frac { 1 }{ 2 } \right) =\frac { \sqrt { \pi } }{ 2 } \]

Similarly, \({ I }_{ 2 }=\sqrt { \pi } \)

Hence, \[\boxed{\frac{{ I }_{ 1 }}{{ I }_{ 2 }}=\frac{1}{2}}\]

\[I=\int_0^\infty x^3 \sum_{n=1}^\infty (-e^{-\alpha x})^n dx\] \[I= \sum_{n=1}^\infty (-1)^n \int_0^\infty x^3 e^{-\alpha n x}dx=\sum_{n=1}^\infty (-1)^n \dfrac{6}{(\alpha n)^4}\] \[I=\dfrac{6}{\alpha^4} (1-2^{-3}) \zeta(4)=\dfrac{1}{64}\] the result follows.

Is the answer \(\huge \displaystyle \alpha = \pi {\left(\frac{56}{15}\right)}^{\frac14}\) ?

Solution to Problem 17:

Maclaurin series of \(\arcsin { \left( x \right) } \) is: \(\displaystyle \arcsin { \left( x \right) } =\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ \sqrt { \pi } \left( 2n+1 \right) n! } { x }^{ 2n+1 } } \)

Here \(x=c \cos(x)\)

\[\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ \sqrt { \pi } \left( 2n+1 \right) n! } { \left( c\cos { \left( x \right) } \right) }^{ 2n+1 } } dx } \]

\[\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) { c }^{ 2n+1 } }{ \sqrt { \pi } \left( 2n+1 \right) n! } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( \cos { \left( x \right) } \right) }^{ 2n+1 }dx } } \]

Therefore, by beta function we get:

\[\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\sum _{ n=0 }^{ \infty }{ \left( \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) { c }^{ 2n+1 } }{ \sqrt { \pi } \left( 2n+1 \right) n! }\frac { 1 }{ 2 } \frac { \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( n+1 \right) }{ \Gamma \left( n+\frac { 3 }{ 2 } \right) } \right) } \]

Hence, on simplifying, we get: \[\boxed{ \displaystyle \int_{0}^{\pi/2} {\arcsin{(c \cos{(x)})} \ dx} = \frac{c}{1^2} + \frac{c^3}{3^2} + \frac{c^5}{5^2} + \cdots \text{ (Upto Infinity)}}\]

Very True.I will change the note after it crosses 50 questions.

I have come up with several ways to prove it. Here is one of the more interesting methods.

Lemma: \[ \displaystyle f(a) = \dfrac{1}{1+a} + \sum_{r=1}^{\infty} \left[ \dfrac{1}{(2r+a+1)}\prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right] = \dfrac{\pi}{2^{a+1}} \dfrac{\Gamma (a+1)}{\Gamma^2\left(\dfrac{a+2}{2}\right)} \quad ; \quad a > -1\]

Proof: I have proved it here

Now,

\(\displaystyle \text{I} = \int_{0}^{\pi /2} {S \ \mathrm{d}x} = \sum_{r=1}^{\infty}\int_{0}^1 \dfrac{\sin ^{2r}}{(2r-1)^2} \mathrm{d}x\)

\(\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\operatorname{B}\left(r+\frac{1}{2} , \frac{1}{2} \right)}{(2r-1)^2}\)

\(\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right) \Gamma \left(\frac{1}{2} \right)}{(2r-1)^2 \Gamma(r+1)}\)

\(\displaystyle = \dfrac{\pi}{2} \sum_{r=1}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right) }{(2r-1)^2 \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}\)

Changing the index of summation and using \(\Gamma(x+1) = x\Gamma(x)\), we have,

\(\displaystyle \text{I} = \dfrac{\pi}{4} \sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(r+1)(2r+1) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}\)

\(\displaystyle = \dfrac{\pi}{2} \left[\sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(2r+1) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)} - \sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(2r+2) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}\right]\)

Using my solution here, we have,

\(\displaystyle \text{I} = \dfrac{\pi}{2} \left[ 1+ \sum_{r=1}^{\infty} \dfrac{1}{(2r+1)} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) - \dfrac{1}{2} - \sum_{r=1}^{\infty} \dfrac{1}{(2r+2)} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right]\)

\(\displaystyle = \dfrac{\pi}{2} \left[f(0) - f(1) \right]\)

\(\displaystyle = \boxed{\dfrac{\pi ^2}{4} - \dfrac{\pi}{2}}\)

What was your intended solution?

@Ishan Singh @Rajdeep Dhingra Do any of you know another method to solve this question?

Solution to Problem 6:

I'll use the definition: \({ H }_{ x }=\psi \left( x+1 \right) +\gamma \)

\[\int { { H }_{ x }dx } =\int { \left( \psi \left( x+1 \right) +\gamma \right) dx } \]

We know that \(\displaystyle \psi \left( x \right) =\frac { d\left( x! \right) }{ dx } \). Hence, using this, we get: \[\boxed{\displaystyle \int { { H }_{ x }dx } =ln\left( x! \right) +\gamma x+c}\]

Let \(\displaystyle \text{I} = \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx\)

Substitute \(\displaystyle x = \frac{1}{t^n}\)

\(\displaystyle \implies \text{I} = n \int_{1}^{\infty} \dfrac{ \left \{t \right \} }{t^{n+1}} dt \)

\(\displaystyle = n \sum_{r=1}^{\infty} \int_{r}^{r+1} \dfrac{t-r}{t^{n+1}} dt\)

\(\displaystyle = n \sum_{r=1}^{\infty} \left[ \dfrac{t^{1-n}}{1-n} +r \dfrac{t^{-n}}{n} \right]_{r}^{r+1} \)

\(\displaystyle = -n \sum_{r=1}^{\infty} \left( \dfrac{1}{n(n+1) (r+1)^{n-1}} + \dfrac{1}{n(r+1)^n} \right) \)

\(\displaystyle = \dfrac{n}{n-1} - \zeta(n) \)

Does the Curly brackets mean fractional part ?

Solution to Problem 9

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Solution to Problem 25:

\[I=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } } \]

\[I=\int _{ 0 }^{ 1 }{ \left( \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dy } \right) dx } \]

By IBP, we get: \[I={ \left( \frac { { x }^{ k+1 } }{ k+1 } \int _{ { x }^{ k } }^{ \infty }{ \frac { \left\{ t \right\} }{ { t }^{ 2 } } dt } \right) }_{ 0 }^{ 1 }+\frac { k }{ k+1 } \int _{ 0 }^{ 1 }{ { x }^{ k }dx } \]

Hence on simplifying, we get: \[\displaystyle\int _{0}^{1}{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } =\frac { 2k+1 }{ (k+1)^{ 2 } } -\frac { \gamma }{ k+1 }} \]

Solution to Problem 22:

Use the formula \(2\sin { \left( A \right) } \cos { \left( B \right) } =\sin { \left( A+B \right) } -\sin { \left( A-B \right) } \)

Therefore, we get: \[ I=\frac { 1 }{ 2 } \left\{ \int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } +\int _{ 0 }^{ \infty }{ \frac { \sin { \left( \frac { 1 }{ x } \right) } }{ x } dx } \right\} \]

It is easy to see that both the integrals are same. Hence, \[I=\int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } =\frac{\pi}{2}\]

solution to problem 8

Considering the function \(f(t) = \int_{0}^{2\pi} e^{t \cos(\theta)} \cos(t\sin(\theta)) \, d\theta\), we see that \[ \begin{array}{rcl} tf'(t) & = & \displaystyle t \int_0^{2\pi} e^{t\cos\theta}\big[\cos\theta \cos(t\sin\theta) - \sin\theta\sin(t\sin\theta)\big]\,d\theta \\ & = & \displaystyle \int_0^{2\pi} \frac{\partial}{\partial \theta}\big[e^{t\cos\theta} \sin(t\sin\theta)\big]\,d\theta \\ & = & \Big[ e^{t\cos\theta}\sin(t\sin\theta)\Big]_0^{2\pi} \; = \; 0\\ &&f'(t)=0\to f(t)=C \end{array} \] Since \(f(0) = 2\pi\), the desired integral is \(f(1) = \boxed{2\pi}\).

Solution to Problem 3

Let us start with the integral \[I(a) = \int_{0}^{1}{ \dfrac{\ln{(1+ax)}}{x\sqrt{1-x^2}} \ dx}\]. Differentiate with respect to a. \[I'(a) = \int_{0}^{1}{ \dfrac1{(1+ax)\sqrt{1-x^2}} \ dx}\] This can easily be solved by substituting \(1 + ax = \dfrac1t\).We now get \[I'(a) = \dfrac{\arccos{(a)}}{\sqrt{1-a^2}}\] Now we find I(a). \[I(a) = \int {\dfrac{\arccos{(a)}}{\sqrt{1-a^2}} \ da} = -\dfrac12\arccos^2{(x)} + C\] Now we put a = 1. We get \[I(1) = C\] Now we put a = -1. We get \[I(-1) = -\dfrac{{\pi}^2}2 + C\] We get \[I(1) - I(-1) = \dfrac{{\pi}^2}2\]

Q.E.D

Lemma : \[\displaystyle \int_{0}^{\infty} \dfrac{\cos x - e^x}{x} \mathrm{d}x = 0\]

Proof : \(\displaystyle \int_{0}^{\infty} \dfrac{\cos x - e^x}{x} \mathrm{d}x = \int_{0}^{\infty} \int_{0}^{\infty} \left(e^{-yx}\cos x - e^{-x(y+1)} \right) \mathrm{d}x \ \mathrm{d}y \)

\(\displaystyle = \int_{0}^{\infty} \left( \dfrac{y}{y^2 + 1} - \dfrac{1}{y+1} \right) \mathrm{d}y\)

\(\displaystyle = \left[\ln \left( \dfrac{\sqrt{y^2+1}}{y+1} \right)\right]_{y=0}^{y \to \infty} = 0\)

Proposition : \[\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \gamma \]

Proof : \(\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x \)

\(\displaystyle = \int_{0}^{\infty} \int_{0}^{1} (1-e^{-x})e^{-yx} \mathrm{d}x \ \mathrm{d}y - \int_{0}^{\infty} \int_{1}^{\infty} e^{-x(y+1)} \mathrm{d}x \ \mathrm{d}y \)

\(\displaystyle = \int_{0}^{\infty} \left(\dfrac{1-e^{-y}}{y} + \dfrac{e^{-(y+1)} - 1}{y+1} \right) \mathrm{d}y - \int_{0}^{\infty} \dfrac{e^{(y+1)}}{y+1} \mathrm{d}y \)

\(\displaystyle = -\int_{0}^{\infty} \left(e^{-y} - \dfrac{1}{y+1} \right)\dfrac{\mathrm{d}y}{y} \)

\(\displaystyle = -\int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x}\)

Using the Lemma,

\(\displaystyle = - \int_{0}^{\infty} \left(\cos x - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x} \)

\(\displaystyle = \gamma\) (Using my solution here (see Problem 39))

Now,

\(\displaystyle \text{I} = \int_{0}^{\infty} \dfrac{\cos x - e^{-x^2}}{x} \mathrm{d}x \)

\(\displaystyle = \int_{0}^{\infty} \dfrac{e^{-x} - e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Lemma}) \)

\(\displaystyle = - \int_{0}^{\infty} \dfrac{1-e^{-x}}{x} \mathrm{d}x + \int_{0}^{\infty} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x\)

\(\displaystyle = -\int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{1-e^{-x}}{x} \mathrm{d}x + \int_{0}^{1} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x + \int_{1}^{\infty}\dfrac{1-e^{-x^2}}{x} \mathrm{d}x \)

\(\displaystyle = -\gamma + \int_{0}^{1} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Proposition}) \)

Substitute \(x^2 \mapsto x\)

\(\displaystyle \implies \text{I} = -\gamma + \dfrac{1}{2} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \mathrm{d}x \right)\)

\( \displaystyle = -\gamma + \dfrac{\gamma}{2} \quad (\text{Using Proposition})\)

\( \displaystyle = \boxed{-\dfrac{\gamma}{2}}\)

Solution to Problem 24 :

Let \[I = \int_{-\infty}^{\infty} { e^{- x^2} \ dx}\] We can write it as \[I = \sqrt{I^2}\] \[ I = \sqrt{\int_{-\infty}^{\infty} { e^{- y^2} \ dy} \int_{-\infty}^{\infty} { e^{- x^2} \ dx}} \] \[I = \sqrt{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} { e^{- x^2 + y^2} \ dy \ dx }}\] Switching to Polar Coordinates of one integral. \[I = \sqrt{\int_{0}^{2\pi} { \int_{0}^{\infty} { e^{-r^2} r \ dr \ d{\theta}}}} \] The 'dr' one integral can be calculated by integration by parts. The other one is very easy. \[I = \sqrt{ \pi}\]

\(\displaystyle \int_{0}^1 \dfrac{\ln (x^2+x+1)}{x} dx = \int_{0}^1 \dfrac{\ln (1-x^3)}{x} dx - \int_{0}^1 \dfrac{\ln (1-x)}{x} dx\)

\(\displaystyle = -\sum_{n=1}^{\infty} \int_{0}^1\dfrac {x^{3n-1}}{n} dx + \sum_{n=1}^{\infty} \int_{0}^1 \dfrac {x^{n-1}}{n} dx \)

\(\displaystyle = \dfrac {2}{3} \zeta (2)\)

solution to problem 11 \[\int uv'=uv-\int u'v\] \[I=\int_0^{\pi/2} x^2 \csc^2(x) dx\] \[u=x^2\to u'=2x, v'=\csc^2 x\to v=-cot(x)\] \[I=-x^2\cot(x)\mid_0^{\pi/2} +2\int_0^{\pi/2} x \cot(x) dx=2\int_0^{\pi/2} x \cot(x) dx\] \[u=x\to u'=1, v'=\cot(x)\to v=-\ln(sin(x))\] \[I=2x\ln(\sin(x))\mid_0^{\pi/2}+2\int_0^{\pi/2} \ln(\sin(x)) dx=\int_0^{\pi/2} \ln(\sin^2(x)) dx\] \[B(a,1/2)=\int_0^{\pi/2} 2\sin^{2a-1}(x) dx\] \[B'(a,1/2)=\int_0^{\pi/2} 2\ln(\sin^2(x))\sin^{2a-1}(x) dx\] \[B'(a,1/2)=B(a,1/2)(\psi(a)-\psi(a+1/2))\] \[a=1/2\] \[\int_0^{\pi/2} 2\ln(\sin^2(x))dx=B(1/2,1/2)(\psi(1/2)-\psi(1))=\pi(-\gamma-(-\gamma-2\ln(2)))=2\pi\ln(2)\] \[I=\int_0^{\pi/2} \ln(\sin^2(x)) dx=\boxed{\pi\ln(2)}\]

Solution to Problem 5:

\[I=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( 1+x \right) } }{ x } dx } \]

\[I=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( -x \right) } }{ 1-x } dx } \]

\[I=-\int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) } }{ 1+x } dx } \]

Now, I'll use power series of \(\frac{1}{x+1}\), I'll write the integral as:

\[I=-\sum _{ n=0 }^{ \infty }{ { \left( -1 \right) }^{ n }\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { x } dx } } \]

\[I=\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } \]

\[\boxed{\therefore I=\frac { \zeta \left( 2 \right) }{ 2 } =\frac { { \pi }^{ 2 } }{ 12 } }\]

solution to problem 5

replace x with -x to have \[\int_0^{-1} \dfrac{\ln(1-x)}{x} dx\] \[\sum_{n≥1} \dfrac{x^n}{n}=\ln(1-x)\] dividing by x and integrating from 0 to a \[\int_0^{a} \dfrac{\ln(1-x)}{x} dx=\sum_{n≥1} \dfrac{a^n}{n^2}\] We have \[\int_0^{-1} \dfrac{\ln(1-x)}{x} dx= \sum_{n≥1}\dfrac{(-1)^n}{n^2}=\sum_{n≥1} \dfrac{(-1)^n}{n^2}=\sum_{n≥1} \dfrac{1}{n^2}-2\sum_{n≥1} \dfrac{1}{(2n)^2}=\\\dfrac{1}{2}\sum_{n≥1} \dfrac{1}{n^2}=\boxed{\dfrac{\pi^2}{12}}\]

Lemma :

\( \displaystyle \int_{0}^{1} x^{m}\ln^{n}(x)dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}} \)

Consider the integral,
\( \displaystyle \int_{0}^{1} x^{m}dx = \dfrac{1}{m+1} \)

Differentiate n times with respect to m,
\( \displaystyle \int_{0}^{1} x^{m}\ln^{n}x dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}} \)

\( I = \displaystyle \int_{0}^{1} \dfrac{ln^{2}(1-x)}{x}dx = \int_{0}^{1} \dfrac{\ln^{2}(x)}{1-x} dx \)

Using the Taylor series for \( \dfrac{1}{1-x} \)

\( I = \displaystyle \sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln^{2}(x)dx \).
Using the Lemma,
\( I = \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^{2} 2!}{(k+1)^{3}} = \sum_{k=1}^{\infty} \dfrac{2}{k^{3}} = 2\zeta(3) \)

Solution to Problem 16

It is a long tedious one. I will just mention some main steps.

Apply integration by parts with 1st function as \(\arctan{(x)}\) and 2nd one as \(\frac{dx}{x^{11}}\).

Now you will get an integral \(\displaystyle \int{\frac1{10x^{10}(x^2 + 1)} \ dx} \).

Apply partial fractions you will get \[\frac1{x^{10}} - \frac1{x^8} + \frac1{x^6} - \frac1{x^4} + \frac1{x^2} - \frac1{x^2 +1}\] Rest all is trivial Integration.

Is the answer \(\displaystyle \large \sum_{k = 1}^{5}{\dfrac{{(-1)}^k}{10(2k-1){x}^{2k-1}}} - \frac{\arctan{(x)}}{10} - \frac{\arctan{(x)}}{10x^{10}}\) ?

Solution to Problem 13

Let us start with \[I = \int_{0}^{\pi/2} { \log{(\cos{(x)})} \ dx} \] Using properties we know that \[I = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx} \] Adding both we get \[2I = \int_{0}^{\pi/2} { \log{(\sin{(x)}\cos{(x)})} \ dx}\] Using little manipulation we get that \[2I = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} - \int_{0}^{\pi/2} { \log{(2)} \ dx} \] Using lemma we get \[2I = I - \dfrac{\pi}2 \log{(2)} \\ \Rightarrow \boxed{I = - \dfrac{\pi}2 \log{(2)} }\]

Proof of Lemma Used

\[I_1 = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} \] Substitute 2x = t. You will get \[I_1 = \frac12 \int_{0}^{\pi} { \log{(\sin{(t)})} \ dt} \\I_1 = \frac12 \left( \int_{0}^{\pi/2} { \log{(\sin{(t)})} \ dt} + \int_{\pi/2}^{\pi} { \log{(\sin{(x)})} \ dx} \right) \] Now let us substitute \(x - \frac{\pi}2 = t\) in the 2nd integral. Then using properties and manipulating. we get \[I_1 = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx}\]

Delete this question! This question is on brilliant. See this.

Simple derivative of beta function.

\(B(a,b) = \displaystyle \int_{0}^{1} x^{m-1} (1-x)^{n-1}dx \)

Required integral is \(\dfrac{\partial B(m,n)}{\partial m \partial n}\) at \(m=n=2\)

This evaluates to\(\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (((\psi (m)-\psi (m+n))(\psi (n)-\psi (m+n))-\psi '(m+n))\).

On putting the values of m and n, the final answer evaluates to \(0.0684\).

Let us start with \[I = \int_{0}^{1}{ \log{(\Gamma{(x)})}cos(16\pi x) \ dx}\] We can also write \[I = \int_{0}^{1}{ \log{(\Gamma{(1-x)})}cos(16\pi - 16\pi x) \ dx} = \int_{0}^{1}{ \log{(\Gamma{(1-x)})}cos(16\pi x) \ dx}\] Adding both we get \[2I = \int_{0}^{1}{ \log{(\Gamma{(1-x)}\Gamma{(x)})}cos(16\pi x) \ dx}\] Using Euler Reflection Formula we get \[2I = \int_{0}^{1}{ \log{(\pi)}cos(16\pi x) \ dx - \int_{0}^{1}{ \log{(\sin{(\pi x)})}cos(16\pi x) \ dx}}\] Both are standard computations. \[2I = 0 - \left(-\dfrac1{16}\right)\] \[\boxed{I = \frac{1}{32}}\] Q.E.D

Solution to Problem 15:

\[I=\int { \frac { \sin ^{ 9 }{ x } }{ \cos ^{ 8 }{ x } } dx } \]

Substitute: \(\cos x=t\)

\[I=-\int { \left( { t }^{ -8 }-{ 4t }^{ -6 }+{ 6t }^{ -4 }-{ 4t }^{ -2 }+1 \right) } dt\]

\[\therefore I=-\cos { \left( x \right) } +\frac { \sec ^{ 7 }{ \left( x \right) } }{ 7 } -\frac { 4\sec ^{ 5 }{ \left( x \right) } }{ 5 } +2\sec ^{ 3 }{ \left( x \right) } -4\sec { \left( x \right) } \]

\( \cos x = t \)
\( \therefore -\sin x dx = dt \)
\( I = \displaystyle \int \dfrac{(1-t^{2})^{4}}{t^{8}}\cdot -dt \)
\( I =- \displaystyle \int \dfrac{1-4t^{2} + 6t^{4} -4t^{6} + t^{8}}{t^{8}} dt \)
\( I = - \left( \dfrac{-1}{7t^{7}} + \dfrac{4}{5t^{5}} - \dfrac{6}{3t^{3}} + \dfrac{4}{t} + t \right) \) + c \( I = - \left( \dfrac{-1}{7\cos^{7} x} + \dfrac{4}{5\cos^{5} x} - \dfrac{6}{3\cos^{3}x}+ \dfrac{4}{\cos x} + \cos x \right) + c \)