home.social

#definiteintegral — Public Fediverse posts

Live and recent posts from across the Fediverse tagged #definiteintegral, aggregated by home.social.

  1. An excellent general result.

    If \(\Re(s)>1\),
    \[\displaystyle\int_0^\infty\frac{\ln x}{x^s+1}~\mathrm dx=\frac{\pi^2}{4s^2}\left[\sec^2\left(\frac{\pi}{2s}\right)-\csc^2\left(\frac{\pi}{2s}\right)\right]\]

    Special cases:
    \[\displaystyle\int_0^\infty\frac{\ln x}{x^2+1}~\mathrm dx=0\]
    \[\displaystyle\int_0^\infty\frac{\ln x}{x^3+1}~\mathrm dx=-\frac{2\pi^2}{27}\]
    \[\displaystyle\int_0^\infty\frac{\ln x}{x^4+1}~\mathrm dx=-\frac{\pi^2}{8\sqrt2}\]
    \[\displaystyle\int_0^\infty\frac{\ln x}{x^5+1}~\mathrm dx=-\frac{4\pi^2}{25}\left(\frac{2+\sqrt5}{5+\sqrt5}\right)=-\frac{(5+3\sqrt5)\pi^2}{125}\]

    #Integral #Integrals #GeneralResult #GeneralResults #Result #Results #Logarithms #Logarithm #Integration #DefiniteIntegral #Calculus #IntegralCalculus

  2. Take a look at this interesting integral. #MathsChallenge [Hint: use the properties of the Jacobi theta function of the third type \(\vartheta_3(z,q)\)]

    \[\boxed{\displaystyle\int_0^{\frac{\pi}{4}}\dfrac{1+2\displaystyle\sum_{n\geq1}e^{-n^2\pi x}}{1+2\displaystyle\sum_{n\geq1}e^{-n^2\pi/x}}\ \mathrm{d}x=\sqrt\pi}\]

    #MathChallenge #IntegralChallenge #InterestingIntegral #WeirdIntegral #Integral #Integrals #DefiniteIntegral

  3. Interesting integral! #Challenge
    \[\displaystyle\int_0^1\dfrac{\ln (x)\ln(1-x)}{x(1-x)}\operatorname{Li}_2(x)\ dx=5\zeta(2)\zeta(3)-8\zeta(5)\]
    Where \(\operatorname{Li}_2(x)\) denotes the dilogarithm (or Spence's function), and \(\zeta(x)\) denotes the Riemann zeta function.

    #ZetaFunction #Zeta #Dilogarithm #SpenceFunction #Polylogarithm #Integral #DefiniteIntegral #Integration #Integrals #RiemannZetaFunction #Logarithm #Function #LogarithmicFunction

  4. Interesting integral! #Challenge
    \[\displaystyle\int_0^1\dfrac{\ln (x)\ln(1-x)}{x(1-x)}\operatorname{Li}_2(x)\ dx=5\zeta(2)\zeta(3)-8\zeta(5)\]
    Where \(\operatorname{Li}_2(x)\) denotes the dilogarithm (or Spence's function), and \(\zeta(x)\) denotes the Riemann zeta function.

    #ZetaFunction #Zeta #Dilogarithm #SpenceFunction #Polylogarithm #Integral #DefiniteIntegral #Integration #Integrals #RiemannZetaFunction #Logarithm #Function #LogarithmicFunction

  5. How to prove this?!🤔

    \[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]

    where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.

    #Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm

  6. How to prove this?!🤔

    \[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]

    where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.

    #Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm

  7. How to prove this?!🤔

    \[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]

    where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.

    #Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm

  8. How to prove this?!🤔

    \[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]

    where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.

    #Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm

  9. How to prove this?!🤔

    \[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]

    where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.

    #Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm

  10. SOME EXOTIC INTEGRALS:

    \[\displaystyle\int_0^1\dfrac{dx}{x^x}=\sum_{k=1}^\infty\dfrac{1}{k^k}\]
    \[\displaystyle\int_0^1x^x\ dx=\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{k^k}\]
    \[\displaystyle\int_0^1x^{x^2}\ dx=\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{(2k-1)^k}\]
    \[\displaystyle\int_0^1x^{\sqrt{x}}\ dx=\sum_{k=1}^\infty(-1)^{k-1}\left(\dfrac{2}{k+1}\right)^k\]

    \[\boxed{\boxed{\displaystyle\int_0^1x^{cx^a}\ dx=\sum_{k=1}^\infty\dfrac{(-c)^{k-1}}{((k-1)a+1)^k}}}\]
    #integral #calculus #definiteintegral