Scroll to see replies
\displaystyle \begin{aligned} \Righatarrow \frac{\partial}{\partial \alpha} \phi(\alpha) =-\int_0^1 \frac{x^{\alpha} (1-x^{\beta})}{(1-x)} dx = -\int_0^1 x^{\alpha} \sum_{n=1}^{\beta} x^{n-1} \ dx = -\sum_{n=1}^{\beta} \int_0^1 x^{\alpha + n-1} \ dx \end{aligned}[br]\displaystyle =-\sum_{n=1}^{\beta} \left[\frac{x^{\alpha +n}}{\alpha +n} \right]_0^1 = -\sum_{n=1}^{\beta} \frac{1}{\alpha +n}
\displaystyle \begin{aligned} \Rightarrow \mathcal{L} \{ f(t) \} = \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-st} \cos(tx) \ dt \dx = \int_0^{\infty} \frac{s}{x(x^2+s^2)} \ dx = \frac{1}{s} \left[ \ln \left( \frac{x}{ \sqrt{x^2+s^2}} \right) \right]_0^{\infty} \end{aligned}
\displaystyle \begin{aligned} \Rightarrow \mathcal{L} \{ f(t) \} = \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-st} \cos(tx) \ dt \dx = \int_0^{\infty} \frac{s}{x(x^2+s^2)} \ dx = \frac{1}{s} \left[ \ln \left( \frac{x}{ \sqrt{x^2+s^2}} \right) \right]_0^{\infty} \end{aligned}
\beginaligned \displaystyle \frac{ \cos {a_1 x} - \cos{a_2 x}}{x} = \displaystyle \int_{a_2}^{a_1} - \sin {ax}\ da = Im \left[ \displaystyle \int_{a_2}^{a_1} e^{-iax}\ da \right][br][br]\Rightarrow \displaystyle \int_0^{\infty} \displaystyle \frac{ \cos {a_1 x} - \cos{a_2 x}}{x}\ dx = \displaystyle \int_0^{\infty} Im \left[ \displaystyle \int_{a_2}^{a_1} e^{-iax}\ da \right]\ dx[br][br]\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \displaystyle \int_0^{\infty} \left[ e^{-iax} \right]\ dx\ da[br][br][br]\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \left[ - \frac{e^{-iax}}{ia} \right] _0^{\infty}\ da[br][br][br]\beginaligned = \displaystyle \int_{a_2}^{a_1} Im \left[ -\frac{i}{a} \right]\ da[br][br][br]\beginaligned = \displaystyle \int_{a_2}^{a_1} -\frac{1}{a}\ da[br][br][br]\beginaligned = \displaystyle \ln { \frac{a_2}{a_1}}[br]
\displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx[br][br]\beginaligned = A_1 \displaystyle \int_0^{\infty} \displaystyle \frac{ \cos {a_1 x} - \cos{a_k x}}{x}\ dx + \cdots A_{k-1}\displaystyle \int_0^{\infty} \displaystyle \frac{ \cos {a_{k-1} x} - \cos{a_k x}}{x}\ dx
[br][br]\beginaligned = A_1 \ln {\frac{a_k}{a_1}} + \cdots A_{k-1} \ln {\frac{a_k}{a_{k-1}}}
[br][br]\beginaligned = \ln {\frac{a_k^{A_1+A_2 + \cdots}}{a_1^{A_1} a_2^{A_2} \cdots}}
[br][br]\beginaligned = \ln {\frac{a_k^{-A_k}}{a_1^{A_1} a_2^{A_2} \cdots}}
[br][br]\beginaligned = - \ln {(a_1^{A_1} a_2^{A_2} \cdots a_k^{A_k})}[br][br]
Spoiler
Last reply 3 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71
Last reply 3 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71