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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

Note by Jatin Yadav
4 years, 1 month ago

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Short and Sweet! :)

Snehal Shekatkar - 4 years, 1 month ago

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Thanks!

Jatin Yadav - 4 years, 1 month ago

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

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

Ankit Sultana - 3 years, 9 months ago

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Solve his other question too: Problem without words !

Pi Han Goh - 4 years, 1 month ago

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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 - 4 years, 1 month ago

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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 - 4 years, 1 month ago

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Hi, Bhargav, you are close, try to use expansion for \(e^x\).

Jatin Yadav - 4 years, 1 month ago

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Is the answer \(\pi\)?

Bhargav Das - 4 years, 1 month ago

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@Bhargav Das I reach the same answer but is it possible to evaluate \(B\)?

Pranav Arora - 4 years, 1 month ago

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\(W0w\) \(£xcellent approach\)

Harsh Soni - 2 years, 11 months ago

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It is an amazing approach to solve such problems.

Deepak Pant - 3 years, 2 months ago

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

Prabir Chaudhuri - 3 years, 3 months ago

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Finally I got it! :)

Pranjal Jain - 3 years, 3 months ago

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Interesting method!

Paramjit Singh - 3 years, 8 months ago

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Easy approach....nice trick...

Prabhat Kumar - 3 years, 8 months ago

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Thanks a ton..!

Avadhoot Khairnar - 3 years, 9 months ago

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Answer is π

Ankit Sultana - 3 years, 9 months ago

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Too beautiful

Rishabh Chhabda - 3 years, 9 months ago

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great..:-)

Lalit Pathak - 4 years ago

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

Shakir Khatti - 4 years ago

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Awesome!!

Balaji Dodda - 4 years, 1 month ago

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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 - 4 years, 1 month ago

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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 - 4 years, 1 month ago

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

Saloni Gupta - 4 years, 1 month ago

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