# Tag Info

## Hot answers tagged definite-integrals

11

Set $t=\frac{e^x}{4}$ , or $e^x=4t$ , thus $4dt=e^x dx$ as $x\to-\infty$ then $t\to 0$ as $x\to\ln(4)$ then $t\to 1$ we have $$I=\int_{-\infty}^{\ln(4)}\frac{xe^x}{\sqrt{4e^x-e^{2x}}}dx=\int_{0}^{1}\frac{\ln(4t)}{\sqrt{16t-16t^2}}4dt=\int_{0}^{1}\frac{\ln(4)+\ln(t)}{\sqrt{t-t^2}}dt$$ $$I=\ln(4)\int_{0}^{1}\frac{1}{\sqrt{t-t^2}}dt+\int_{0}^{1}\frac{\ln(t)... 7 It's good, if read correctly; you are less likely to make errors with this if you set$$ G(x)=\int_0^xg(t)\,dt $$and write$$ \int_0^af(x)g(x)\,dx= \Bigl[f(x)G(x)\Bigr]_0^a-\int_0^a f'(x)G(x)\,dx $$If you prefer not to use G, you can write$$ \int_0^af(x)g(x)\,dx= \left[f(x)\int_0^x g(t)\,dt\right] - \int_0^a f'(x)\left(\int_0^x g(t)\,dt\right)\,dx $$5 We can use an integral representation of the Dirichlet eta function to show that$$\int_{0}^{\infty} \frac{\tanh^{2}(x)}{x^{2}} \, dx = \int_{0}^{\infty} \left(1-\frac{1}{\cosh^{2}(x)} \right) \frac{dx}{x^{2}} = -56 \, \zeta'(-2) = \frac{14 \, \zeta(3)}{\pi^{2}}. $$An integral representation of the Dirichlet eta function is$$\eta(s) = \frac{1}{\Gamma(s)...

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On the first line you get $$v(t)=\frac{a_0}{\omega}\cdot (-\cos(\omega \cdot t))+C$$ then by putting $t=0$, you have $v(0)=0$ giving \begin{align} &0=\frac{a_0}{\omega}\cdot (-\cos(\omega \cdot 0))+C \\\\&0=-\frac{a_0}{\omega}+C \\\\&C=\frac{a_0}{\omega}. \end{align} Then there is no more contradiction with your second computation.

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