# Tagged Questions

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### Evaluating the definite integral $\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$

How to evaluate the following integral? $$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$
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### How to calculate $\int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4}$?

I would like to calculate $$\int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4}$$ by means of the Residue Theorem. This is what I tried so far: We can define a path $\alpha$ that consists of half a ...
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### Is the integral $\int^{+\infty}_0 \arcsin (x)\, dx$ improper? Why not?

Is this integral $\displaystyle \int\limits^{+\infty}_0 \arcsin (x)\, dx$ improper? We have $+\infty$, but it's not improper. Why?
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### Is this integral improper? If yes - why?

Is this integral improper? If yes - why? $$\int\limits^2_0 \,\frac{1}{x-1} dx$$
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### Is there a formula for this integral

Is there a formula for the following integral? $$I(a,b)=\int_0^1 t^{-3/2}(1-t)^{-1/2}\exp\left(-\frac{a^2}{t}-\frac{b^2}{1-t} \right)dt$$ where $a,b$ are non-zero real numbers.
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### Improper integral and special functions

I'd like to have an expression of the following integral: $$\int_0^{+\infty} \left(\sqrt{1+x^4} - x^2\right) dx$$ in terms of some special functions (but not in the form given by Wolfram Alpha).
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### Prove that $\int_{-\infty}^{\infty} \sin x \, dx = 0$

$$\int_{-\infty}^{\infty} \sin x \, dx$$ When I am doing the proof for this, why do i have to split it into $\int_{-\infty}^a \sin x \, dx + \int_a^\infty \sin x \, dx$? where a is a constant
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### Inverse Laplace Transform. Computing the integral.

This question is related to this one, but I'm hereby taking a different approach. Problem: Solve $\ddot x+\delta\dot x+\omega_0^2x=\gamma\cos\omega t$. Find the stationary points and examine their ...
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### Integrate: $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$

Can this integral be solved with contour integral or by some application of Residue theorem? $$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catlan constant}$$ It has two ...
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### Complex-valued Fourier integral: $\int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx}$

I'm working on the Fourier transform, but I don't know how to evaluate the integral: $$I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx}$$
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### $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$

I need to find a closed-form for the following integral. Please give me some ideas how to approach it: $$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$
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### What is $\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt$?

I want to compute the following integral: $\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt$ with $\alpha, \beta, c$ real constants, and $\alpha>0,\beta=0$. ...
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### Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick

I stumbled upon this short article on last weekend, it introduces an integral trick that exploits differentiation under the integral sign. On its last page, the author, Mr. Anonymous, left several ...
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### improper integral with square roots

Iv'e ran into this improper integral: $$\int_{0}^{1}(\sqrt{x}/\sqrt{1-x^6})dx$$ I've tried to position $t=\sqrt x$ and i didn't get very close. Any help will be greatly appreciated. thank you, ...
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### Improper Integral:$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$

I want show that this improper integral convergence: $$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$$ please help me.
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### Calculating $\int_{-\infty}^\infty e^{-ax^2}e^{ibx}dx$

In my syllabus about quantum mechanics, they state that the following integral can be easily calculated: $$\int_{-\infty}^\infty e^{-ax^2}e^{ibx}dx = \sqrt{\frac{\pi}{a}}e^{-b^2/4a}$$ if it is ...
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### Calculus - improper integrals

I have a few questions from my h.w, I hope someone can help me. the question is: $f:[a,\infty) \rightarrow \mathbb{R}$ is a continuous and periodic function, with period of $T>0$ . ...
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### Prove the next integral converges for any $x>0$: $\int_0^{\infty}t^{x-1}e^{-t}dt$

Prove the next integral converges for any $x>0$: $$\int_0^{\infty}t^{x-1}e^{-t}dt$$ I can't find a proper way to prove that, But what i did so far was: integration by parts: ...
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### Prove the following equation: $\int_0^{\infty} \frac{\cos{(x)}}{1+x} \,\mathrm dx=\int_0^{\infty} \frac{\sin{(x)}}{(1+x)^2} \,\mathrm dx$

Prove the following equation: $$\int_0^{\infty} \frac{\cos{(x)}}{1+x} \,\mathrm dx=\int_0^{\infty} \frac{\sin{(x)}}{(1+x)^2} \,\mathrm dx$$
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### Determine if the integral converges: $\int_1^{\infty} \frac{\arctan (px)}{x^q}dx$

Determine if the integral converges: $$\int_1^{\infty} \frac{\arctan (px)}{x^q}dx$$ where $p,q\in\Bbb R$.
### Determine if the integral converges: $\int_{-\infty}^{2} \frac{e^{3x}dx}{1+x^2}$
Determine if the integral converges: $$\int_{-\infty}^{2} \frac{e^{3x}dx}{1+x^2}$$ I've tried this: $$f(x)=\frac{e^{3x}}{1+x^2}>\frac{e^{\ln{3x}}}{1+x^2}=\frac{3x}{1+x^2}=g(x)$$ now since g(x) ...