Suppose one integrates $$ \int^1_0 \frac{1}{4y – 1} \, \mathrm{d} y $$ Evaluating one obtains: $$-\infty – 0 + \infty + \frac{\ln 3}{4} $$ which is indeterminate. However, $$\lim_{y \rightarrow \frac{1}{4}} \frac{\ln (4y – 1)}{4} – 0 – \frac{\ln (4y – 1)}{4} + \frac{\ln 3}{4} =\frac{\ln 3}{4} $$ I think this is the Cauchy principal […]

I am trying to compute the function: $$f(\lambda)\equiv\int_{-1}^{1}\frac{\sqrt{1-x^2}}{\lambda-x}dx.$$ It arises as the convolution of the semi-circle density with the inverse function. When $\lambda\in(-1,1)$ it can only be defined as a Cauchy Principal Value. I have a hunch that I need to go into the complex plane to solve this, but am not sure how to […]

How to find the Cauchy principal value of the following integral $$\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx$$ How to start this problem?

I am kinda stuck and not sure what to do at this point of the calculation where: $$\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\,\sqrt{\, 2\,}\,\,\right)^{s}\Gamma\left(\,{s \over 2}\,\right)\,{\rm d}s $$ The Gamma Function produces multiple singularities and I am not sure if the Residue Theorem can be applicable here.

I am at this point of integration where: $$\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left({\sigma\,x^{-1}}\right)^s{\Gamma(\beta_1-1+s)\over \Gamma(\beta_1+\beta_2-1+s)}\,ds$$ whereby $\beta_1$, $\beta_2$, $\sigma$ and $x$ are real and $c>0$ Cauchy’s residue theorem is used and I am not sure how the integral can be simplified to apply the theorem

This is a similar problem to the one I posted here. I am at this point of integration where: $$\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left({\sigma \over x}\right)^s{{1-\beta^{s+1}}\over s(s+1)}\,ds$$ whereby $\beta > 0$, $\sigma > 0$, $c>0$ and $|x|<\beta$. $\beta$, $\sigma$, $c$ and $x$ are all real numbers

I want to compute the following integral $$\oint_{|z|=1}\frac{\exp \left (\frac{1}{z} \right)}{z^2-1}\,dz$$ The integrand has essential singularity at the origin, and $2$-poles at $\pm 1$,which lie on the curve $|z|=1$ so I can’t apply residue formula. How can I proceed?

I am proving this integral: $$ \int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\beta^{1 \over 2}\,\right)^{s}\ \Gamma\left(\,s \over 2\,\right) \Gamma\left(\,{\beta +1 \over 2} – {s \over 2}\,\right)\,{\rm d}s = \Gamma\left(\,{\beta +1 \over 2}\,\right)\left(\,1+{1 \over \beta}\left(x \over \sigma\right)^2\,\right)^{-{\beta +1 \over 2}}$$ where $\beta>0$, $\sigma>0$ and $x$ is real number The clue I have is that Cauchy’s residue theorem […]

I am trying to integrate this integral: $$f(x)=\frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}x^{-s}\sigma ^{ms-m}\left [ \frac{\Gamma \left ( \frac{s}{\beta} \right )}{\Gamma \left ( \frac{1}{\beta} \right )} \right ]^{m}ds$$ where $\sigma>0$, $\beta>0$, $m>0$ and both $\beta$ and $m$ are positive integers For $m=2$, I need help continuing after simplifying to this form below: $$f(x)=\frac{1}{2\pi \sigma^{2} j \Gamma \left ( \frac{1}{\beta} […]

This is the integration I am trying to solve $$\int_{0}^{\pi} \sin^{2}(\theta)\sec^{3}(\theta)d\theta$$ putting $$z=e^{i\theta}$$ $$\int_{\gamma} \frac{-2{(z^{2}-1)}^2}{i(z-i)^{3}(z+i)^{3}}d\theta$$ when applying the residue theorem over a circle of radius 1, singularities are on the circle instead of inside the circle.How can we evaluate a integration like this Thanks

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