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Zero is the correct result. Switching the integrals you have $$\int_1^{+\infty}\frac{\text{d}t}{t^3}\int_0^{+2\pi} \sin(s-t)\ \text{d}s = \int_1^{+\infty}\frac{\text{d}t}{t^3}\left(-\cos(s-t)\bigg|_ …

Notice that $\cos(\pi(n+1)) = -\cos(\pi n)$. Don't know if this could help but as $x\to +\infty$ we have an asymptotic estimation for the integral: $$\int_1^x e^{x/n} \cos(\pi n)\ \text{d}n \sim -x \t …

This is a rather good challenge for numerical analysis. The good thing is that the parameter $a$ belongs in $\mathbb{R}^+$. Well, a "not so basic" knowledge of Special Bessel Functions can help in g …

After point 1) integrate by parts using $$f'(x) = \frac{5^x}{5^x + 1} ~~~~~~~~~~~ \to ~~~~~ f(x) = \frac{\ln(5^x + 1)}{\ln(5)}$$ $$g(x) = x^2 ~~~~~~~~~~~ \to ~~~~~ g'(x) = 2x$$ Then again by parts …

Prologue I won't fix the extrema of the integral after my substitution, and I'll leave you the renaming part because it's simple HINT From where you got stuck, substitute $$t = \frac{1}{\sqrt{2}}\ …

I evaluated it and the answer is $$\frac{a}{2}$$

The integral has no elementary solution, and it's a problem also because to the unknown nature of the limits $a$ and $b$. One way to attack the problem, under some assumptions, is the following: Firs …

The latter series is quite easy only when $p+1 = 0$, which doesn't belong to your case, or when $p+1 = 1$, which might be your case depending on how you define the set $\mathbb{N}$, which usually does …

Let's have a look. First, binomial expansion: $$(1 - a\sin^2(x'))^n = \sum_{k =0}^n (-a \sin^2 x')^k$$ Second, let's expand the sine of the difference $$\sin(x-x') = \sin x \cos x' - \cos x \sin x' …

Hints. $$\int\left(f(x) + f''(x)\right) \sin(x)\ dx = \int f(x)\sin(x)\ dx + \int f''(x)\sin(x)\ dx$$ $$\int f(x)\sin(x)\ dx = \int u(x) v'(x) = u(x) v(x) - \int u'(x) v(x)\ dx$$ With $u = f(x)$, $ …

Just for the love of math, I'll provide you an alternative derivation of that result, by series. $$\int\frac{e^x}{1 + 2e^x}\ \text{d}x = \int\frac{e^x}{e^{2x}(1 + e^{-2x})}\ \text{d}x$$ Given that, …

$a, b$ are positive and the range of integration is $[0, 1]$. What about a Taylor series for the Exponential? $$e^{\left(-\frac{a}{t} - \frac{b}{1-t}\right)} = e^{\frac{t(a-b) -a}{t(1-t)}}$$ From th …

Through Feynman trick we can write $$\int_{-\infty}^{+\infty} -\frac{d}{db} e^{-bx^2}\ dx = -\frac{d}{db}\int_{-\infty}^{+\infty} e^{-bx^2}\ dx$$ The latter integral has to be known, which is $-\sqr …

Let's call for simplicity $y = x/2$ so $\text{d}y = \text{d}x/2$ and $x = 2y$. Then you have $$\int_0^{+\infty} \frac{(2y)^2}{e^y + e^{-y}} 2\ \text{d}y$$ namely $$8\int_0^{+\infty} \frac{(y)^2}{e^ …

Interesting integral. For the moment, let's just rewrite it in a different way by simply unifying the denominator: $$\int_0^k \frac{x^2}{\sqrt{\frac{k^2 - x^2}{k^2}}} \ln\left(1 - e^{-x/t}\right)\ \ …