Q.Evaluate the following integrals:

Concept and Intuition

When an integral over $(0,\infty)$ has an integrand that is "anti-symmetric" under the reciprocal map $x\to1/x$ (i.e., transforms into its own negative), the integral must vanish — this is a classic trick avoiding the need to actually evaluate divergent-looking pieces.

Step-by-Step Solution

1. Let $h(x) = (x^{12}+x^{-12})\dfrac{\log x}{x}$ and $I=\int_0^\infty h(x)\,dx$.
2. Substitute $x=\dfrac1t$, so $dx=-\dfrac{1}{t^2}dt$; as $x:0\to\infty$, $t:\infty\to0$.
3. $x^{12}+x^{-12} = t^{-12}+t^{12}$ (symmetric, unchanged in form). $\log x = -\log t$. $\dfrac{dx}{x} = \dfrac{-dt/t^2}{1/t} = -\dfrac{dt}{t}$.
4. So $h(x)\,dx = (t^{12}+t^{-12})(-\log t)\left(-\dfrac{dt}{t}\right) = (t^{12}+t^{-12})\dfrac{\log t}{t}\,dt$... but we must also flip the limits of integration back from $(\infty,0)$ to $(0,\infty)$, which introduces one more overall minus sign.
5. Carrying the limit-flip sign through carefully: $I = -\displaystyle\int_0^\infty (t^{12}+t^{-12})\dfrac{\log t}{t}dt = -I$.
6. So $2I=0\Rightarrow I=0$. …

Concept and Intuition

$N$'s integrand $\frac{e^{2t}t}{1+e^{3t}}$ is exactly what you get from $\frac{s\log s}{1+s^3}$ under $s=e^t$ — recognizing this lets us relate $N$ back to an integral of the same shape as $M$, and then a reciprocal substitution inside $M$ finishes the comparison.

Step-by-Step Solution

1. In $N=\displaystyle\int_{-\infty}^\infty\frac{e^{2t}t}{1+e^{3t}}dt$, substitute $s=e^t$ (so $t=\log s$, $dt=ds/s$, and as $t:-\infty\to\infty$, $s:0\to\infty$):

2. Call this integral $P=\displaystyle\int_0^\infty\frac{s\log s}{1+s^3}ds$, so $N=P$.
3. Now relate $P$ to $M=\displaystyle\int_0^\infty\frac{\log t}{1+t^3}dt$ by substituting $s=1/u$ in $P$ ($ds=-du/u^2$; as $s:0\to\infty$, $u:\infty\to0$): …

Concept and Intuition

Integrands of the form $\dfrac{1}{(x\tan x+1)^2}$ often come from quotient-rule differentiation of $\dfrac{\sin x}{\cos x + x\sin x}$ — spotting this pattern avoids a messy substitution.

Step-by-Step Solution

1. Rewrite $1+x\tan x=\dfrac{\cos x+x\sin x}{\cos x}$, so the integrand is $\dfrac{\cos^2x}{(\cos x+x\sin x)^2}$.
2. Let $D(x)=\cos x+x\sin x$. Then $D'(x)=-\sin x+\sin x+x\cos x=x\cos x$.
3. Consider $g(x)=\dfrac{\sin x}{D(x)}$. Then

4. This exactly matches the integrand! So $f(x)=g(x)=\dfrac{\sin x}{\cos x+x\sin x}$ (up to the constant $c$). …

Concept and Intuition

Whenever you see $\frac1n\sum f(k/n)$ (or an algebraically disguised version of it) as $n\to\infty$, it is begging to be converted into a definite integral $\int_0^1 f(t)\,dt$. The trick here is to first pull the $\sqrt n$ inside the sum so every term becomes a function of $k/n$ times $1/n$.

Step-by-Step Solution

1. Write the expression as $S_n = \dfrac{1}{\sqrt n}\sum_{k=1}^n \dfrac{1}{\sqrt k}$.
2. Multiply and divide inside: $\dfrac{1}{\sqrt n}\cdot\dfrac1{\sqrt k} = \dfrac1n\cdot\sqrt{\dfrac nk} = \dfrac1n\cdot\dfrac{1}{\sqrt{k/n}}$.
3. So $S_n = \displaystyle\sum_{k=1}^n \frac1n\cdot\frac{1}{\sqrt{k/n}}$, which is exactly the Riemann sum (right-endpoint) for $f(t)=1/\sqrt t$ on $[0,1]$ with $t_k = k/n$. …

Concept and Intuition

The sum $1^p+2^p+\dots+n^p$ grows like $\displaystyle\int_0^n x^p\,dx = \dfrac{n^{p+1}}{p+1}$ for large $n$ — this can be seen directly by writing the ratio as a Riemann sum: $\dfrac{1}{n}\sum_{k=1}^n\left(\dfrac kn\right)^p \to \int_0^1 x^p\,dx=\dfrac{1}{p+1}$.

Step-by-Step Solution

1. Write $\dfrac{1^{77}+2^{77}+\cdots+n^{77}}{n^{78}}=\dfrac1n\sum_{k=1}^n\left(\dfrac kn\right)^{77}$.
2. As $n\to\infty$, this is a Riemann sum for $\displaystyle\int_0^1 x^{77}\,dx$.
3. $\displaystyle\int_0^1 x^{77}dx=\left[\dfrac{x^{78}}{78}\right]_0^1=\dfrac{1}{78}$. …