Use of complex numbers in calculus.

Hi, you might have come across: $\displaystyle \int e^{ax}\cos bx$ dx,

How do you solve it? You might use integration by parts, and also complex numbers, and I find use of complex numbers interesting!

Say $A = \displaystyle \int e^{ax}\cos bx$ dx,

and $B = \displaystyle \int e^{ax}\sin bx$ dx

Hence, $A + iB = \displaystyle \int e^{ax} (\cos bx + i \sin bx)$ dx = $\displaystyle \int e^{ax} (e^{i bx})$ dx

= $\displaystyle \int e^{(a+ib)x}$ dx

= $\displaystyle \frac{e^{(a+ib)x}}{a+ib}$

= $\displaystyle \frac{e^{(a+ib)x}(a - ib)}{a^2 + b^2}$

$\Rightarrow A + iB = z = \displaystyle \frac{e^{ax} (\cos bx + i \sin bx)(a - ib)}{a^2+b^2}$

Clearly,

$A = \text{Re}(z) = \displaystyle \frac{e^{ax}}{a^2+b^2} (a \cos bx + b \sin bx)$

$B = \text{Im}(z) = \displaystyle \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)$

You can try to find this definite integral:

Problem: $\displaystyle \int_{0}^{\pi} e^{(\cos x)} \cos(\sin x)$ dx

#Calculus #ComplexNumbers #Goldbach'sConjurersGroup

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Short and Sweet! :)

Snehal Shekatkar - 7 years, 5 months ago

Thanks!

jatin yadav - 7 years, 5 months ago

$cos { (\sin { x) } } =\quad ({ e }^{ isinx }+{ e }^{ -isinx })/2\\ { e }^{ cosx }\cos { (\sin { x) } } =\quad ({ e }^{ cosx+isinx }+{ e }^{ cosx-isinx })/2=({ e }^{ { e }^{ ix } }+{ e }^{ { e }^{ -ix } })/2\\ =((1+\frac { { e }^{ ix } }{ 1! } +\frac { { e }^{ i2x } }{ 2! } +...)+(1+\frac { { e }^{ -ix } }{ 1! } +\frac { { e }^{ -2ix } }{ 2! } +...))/2$

In 2 more steps you will get the answer. The answer in this case is $pi$

Ankit Sultana - 7 years, 1 month ago

Solve his other question too: Problem without words !

Pi Han Goh - 7 years, 5 months ago

Hi Jatin! Thanks for the great post, but just one thing I was wondering that incorporating i, the complex number, and using the usual laws of Calculus, is it mathematically correct, I mean the laws of calculus is for reals. I may be fundamentally wrong somewhere but I need the answer.

Jit Ganguly - 7 years, 5 months ago

Awesome article.

By using your method I arrive at a step from where I can't proceed further.

The step is: $A+iB=\int (e)^{e^{ix}}\,dx$ , where $A=\int (e)^{(cos x)}cos(sin x)\,dx$ and $B=\int (e)^{(cos x)}sin(sin x)\,dx$ .

How to do after this?Precisely,what's the real part?

Bhargav Das - 7 years, 5 months ago

Hi, Bhargav, you are close, try to use expansion for $e^x$ .

jatin yadav - 7 years, 5 months ago

Is the answer $\pi$ ?

Bhargav Das - 7 years, 5 months ago

@Bhargav Das – I reach the same answer but is it possible to evaluate $B$ ?

Pranav Arora - 7 years, 5 months ago

You came our with a simple and an interesting solution the technique that triggered me was to solve it using By Parts Method but have to admit your approach was far far better than mines Can you tell me how do you get such ideas at and tender age of 15(Just Asking)?

Harshil Patel - 7 years, 5 months ago

Hi, solving this integral using complex numbers is well(not very much well though) known. I did not come up with it myself.

jatin yadav - 7 years, 5 months ago

Yeah, actually the same technique's been discussed with us at our insti as well... anyways, it's really good...

Saloni Gupta - 7 years, 5 months ago

Awesome!!

Balaji Dodda - 7 years, 5 months ago

-e^{-1} -1 is the ans

Shakir Khatti - 7 years, 4 months ago

great..:-)

Lalit Pathak - 7 years, 4 months ago

Too beautiful

Rishabh Chhabda - 7 years, 1 month ago

Answer is π

Ankit Sultana - 7 years, 1 month ago

Thanks a ton..!

A Former Brilliant Member - 7 years, 1 month ago

Easy approach....nice trick...

Prabhat kumar - 7 years ago

Interesting method!

A Brilliant Member - 7 years ago

Finally I got it! :)

Pranjal Jain - 6 years, 7 months ago

THANKS. VERY POWERFULL METHOD-CAN BE TACKLED ANY COMPLICATED INTEGRALS LIKE THIS ONE.

Prabir Chaudhuri - 6 years, 6 months ago

It is an amazing approach to solve such problems.

Deepak Pant - 6 years, 6 months ago

$W0w$ $£xcellent approach$

harsh soni - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$