# Tagged Questions

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### Behavior of lower incomplete gamma function at complex infinity

The lower incomplete gamma function is given by $\gamma\left(s, x\right) = \int\limits_0^x t^{s-1} e^{-t} {\rm d} t~,$ and has a well-defined analytic continuation for both $s$ and $x$ [1]. ...
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### Integral of $\ln(x)\operatorname{sech}(x)$

How can I prove that: ...
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### Expressing $\int_{-\infty}^\infty dx/(x^2+1)^n$ in terms of Gamma function

How to prove this identity for $n>1/2$? $$\int_{-\infty}^{\infty}\frac{dx}{(x^2+1)^n}=\frac{\sqrt{\pi}\cdot \Gamma(n-\frac{1}{2}) }{\Gamma (n)}$$
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### Solving integral that contain exponential function and lower incomplete gamma function

I have the following integral; $$y=\int_0^\infty\frac{e^{-xf}}{m+x}\gamma(a,hx)~dx$$ where $f,m,h\in\mathbb{R}^+$ , $a\in\mathbb{N}$ , $\gamma\left(a,h x\right)$ is the lower incomplete gamma ...
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### Compute $\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz$ as a function of $\Gamma(n)$

Is it possible to compute this integral $$\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz$$ as a function of complete gamma $\Gamma(n)$. If possible, I'm looking for a closed form solution. Thanks!
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### Evaluating improper integral expression

Can anybody please guide me in evaluating this expression, for my research work. I have tried a lot but in vain. Both the integrals involve gamma functions and even wolfram said that the time limit ...
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### Evaluating integral using Beta and Gamma functions.

$$\int_0^{\infty} x^{-3/2} (1 - e^{-x})\, dx$$ Evaluate the above integral with the help of Beta and Gamma functions. I'm badly stuck. I'm getting Gamma of a negative number and have no clue how ...
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### How to compute $\int^{\infty}_{0} t^{(\frac1n-1)}\cos t \,\mathrm{d}t$?

How to calculate the below integral? $$\int^{\infty}_{0} \frac{\cos t}{t^{1-\frac{1}{n}}} \textrm{d}t = \frac{\pi}{2\sin(\frac{\pi}{2n})\Gamma(1-\frac{1}{n})}$$ where $n\in \mathbb{N}$.
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### Closed form for $\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}$

Consider the following integral: $$\mathcal{I}=\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}.$$ It can be represented as ...
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### Calculating an almost Gamma integral

How would you proof that $$I:=\int_{0}^{\infty}\frac{z^{x-1}}{e^{z}+1}dz=\left(1-2^{1-x}\right)\Gamma(x)\zeta(x)$$ I can rewrite the integral as ...
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### A $\log$ integral with a parameter

Prove that: $$\int_0^\infty \frac{\ln x}{x^a+1}\;\text{d}x=-\left( \frac{\pi }{a} \right)\cot \left( \frac{\pi }{a} \right)\csc \left( \frac{\pi }{a} \right),\ \ a>1$$ For this one I consider to ...
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Does anyone know how to approximate this infinite integral $$H(\beta, i, l) = \iint_{1 \leq x \leq y < \infty} (1-x^{-\beta})^{i-1} (x^{-\beta} - y^{-\beta})^{2(l-i)} x^{-\beta} y^{-\beta i} dxdy, ... 1answer 136 views ### Convergence of this integral [duplicate] Possible Duplicate: Some questions about the gamma function My statistics text book prescribed by my school states that the integral$$\Gamma(n)=\int_{0}^{\infty}e^{-x}x^{n-1}dx is ...
I know that the Gamma function with argument $(-\frac{1}{ 2})$ -- in other words $\Gamma(-\frac{1}{2})$ is equal to $-2\pi^{1/2}$. However, the definition of $\Gamma(k)=\int_0^\infty t^{k-1}e^{-t}dt$ ...
### Finding $\int_{-\infty}^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx$ by symmetry
I can easily show that (substituting $\frac{x^2}{2} = t$ using the identity for Gamma function of $n+\frac{1}{2}$, then further expanding $\Gamma(n+\frac{1}{2})=\dfrac{(2n-1)!! \sqrt{\pi}}{2^n}$ and ...