1
$\begingroup$

Question:

$$\int \limits _1^4\left (\frac{1}{2t}+i \right )^2\,dt.$$

How can I solve this? I believe its an indefinite integral and I can probably expand it by using $(a+b)^2=a^2+2ab+b^2$ to get something like$$\int \frac{1}{4t^2}+\frac{2i}{2t}+i^2\,dt$$but I'm not even sure if that's right.

$\endgroup$
1
  • 1
    $\begingroup$ That $i$ it's a bad notation here. The $\int_{1}^{4}f(t)\,{\rm d}t$ is a definite integral and yes, you can use $(a+b)^{2}=a^{2}+2ab+b^{2}$. Then use $\int \frac{1}{t}\, {\rm d}t=\ln|t|+c$ and $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$. $\endgroup$
    – Event Horizon
    May 17 at 16:32

2 Answers 2

Reset to default
2
$\begingroup$

Since $$ \left(\frac1{2t}+i\right)^2=\frac1{4t^2}-1+\frac it, $$ you have $$ \begin{align*} \int_1^4\left(\frac1{2t}+i\right)^2\,dt&=\int_1^4\frac1{4t^2}-1\,dt+\left(\int_1^4\frac1t\,dt\right)i\\ &=-\frac{45}{16}+i\log(4). \end{align*} $$

$\endgroup$
2
$\begingroup$

We have \begin{align*}\int \limits _1^4\left (\frac{1}{2t}+i\right )^2\,dt & =\int \limits _1^4\left (\frac{i}{t}+\frac{1}{4t^2}-1\right )\,dt \\ & =i\int \limits _1^4\frac{1}{t}\,dt+\frac{1}{4}\int \limits _1^4\frac{1}{t^2}\,dt-\int \limits _1^41\,dt \\ & =\left .i\ln |t|-t-\frac{1}{4t}\right |_1^4 \\ & =\frac{16i\ln 4-65}{16}+\frac{5}{4} \\ & =\frac{16i\ln 4-45}{16}. \end{align*}

$\endgroup$
3
  • $\begingroup$ If $i^{2}=-1$ so what's the problem? There's not problem. However, I agree that to use $i$ is a bad notation here. $\endgroup$
    – Event Horizon
    May 17 at 16:34
  • $\begingroup$ @EventHorizon no serious problem but as you said, horrible notation. $\endgroup$
    – algevristis
    May 17 at 16:35
  • $\begingroup$ @EventHorizon edit: I automatically assumed that $i^2 = -1$ anyways to calculate it... $\endgroup$
    – algevristis
    May 17 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.