#DefiniteIntegral

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-07-20

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

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-05-13

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

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-05-12

Integral challenge!

\[\displaystyle\int_0^\infty\ln\left(1+\dfrac{\cosh\alpha}{\cosh x}\right)\ dx=\dfrac{\pi^2}{8}-\dfrac{\arccos^2(\cosh\alpha)}{2}\]

#Integral #Integrals #IntegralChallenge #HyperbolicFunction #HyperbolicCosine #Logarithm #DefiniteIntegral

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2024-05-12

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

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2023-02-24

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

Pustam | पुस्तम | পুস্তম🇳🇵pustam_egr@mathstodon.xyz
2022-12-31

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

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